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unit 17 parabola with solution

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unit 17 parabola with solution

- 1. PARABOLA A conic section is a section cut off from a right circular cone by a plane in various ways. The shape of the section depends upon the position of the cutting plane. 1. If the plane passes through the axis of the cone, the curve of intersection will be a pair of straight lines. 2. If the plane is perpendicular to the axis of the cone, the curve of intersection will be a circle. 3. If the plane is parallel to a generator of the cone (Say, PQ), the curve of intersection will be a parabola. the cone at an angle (0 4. If the plane cuts the axis of < < /2), the curve of intersection will be an ellipse. V A P Q B O Plane P Q O V Generator Parabola Plane P Q O V Ellipse Plane P Q O V Plane Circle Circular base
- 2. 5. If the plane is parallel to the axis of the cone, the curve of intersection will be a hyperbola. 1.1 DEFINITION OF CONIC A conic section or conic is the locus of a point, which moves such that its distance from a fixed point is in a constant ratio to its distance from a fixed straight line, not passing through the fixed point. (i) The fixed point is called the focus. (ii) The fixed straight line is called the directrix. (iii) The constant ratio is called the eccentricity of the conic and is denoted by e. (iv) When the eccentricity is unity; i.e., e = 1, the conic is called a parabola; when e < 1, the conic is called an ellipse; and when e > 1, the conic is called a hyperbola. (v) The line of symmetry of the conic section is called its axis. (vi) A point of intersection of a conic with its axis is called vertex. (vii) The chord of a conic which passes through the focus and perpendicular to the axis is called the latus rectum. 1.2 GENERAL EQUATION OF CONIC Let S (, ) be the focus and Ax + By + C = 0 be the equation of the directrix QN of the conic section. Let P (x, y) be any point on it and let PN QN. If ‘e’ be the eccentricity of the conic, then by definition, PN PS = e PS2 = e2 . PN2 ……(1) (x – )2 + (y – )2 = e2 2 2 2 B A C By Ax which on simplification takes the form Q N P (x, y) S (, ) ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ……(2) where a, b, c, f, g and h are constants. (2) being the locus of P, is the equation of the conic. 1.3 RECOGNITION OF CONICS The general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents A pair of straight line If = 0 where = abc + 2fgh - af2 - bg2 - ch2 , 2 h ab Hyperbola Axis
- 3. A circle if 0, h = 0, a = b A parabola if 0, ab - h2 = 0 An ellipse if 0, ab - h2 > 0 An hyperbolaif 0, ab – h2 < 0 Example -1: Find the equation of the parabola whose focus is (3, -4) and directrix is x - y + 5 = 0. Solution: Let P(x, y) be any point on the parabola. Then 1 1 5 y x 4 y 3 x 2 2 (x - 3)2 + (y + 4)2 = 2 5 y x 2 x2 + y2 + 2xy - 22x + 26y + 25 = 0 (x + y)2 = 22x - 26y – 25. Example -2: If (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then its equation is (A) x2 + 8y = 32 (B) y2 + 8x = 32 (C) x2 - 8y = 32 (D) y2 - 8x = 32 Solution: AS = 2 = a. Vertex (0, 4) lies on y-axis. Hence the parabola X2 = -4aY is a downward parabola as focus is below the vertex. Or (x –0)2 = -4 2(y –4) Or x2 + 8y = 32 Hence (A) is the correct answer. A (0, 4) S (0, 2) X Y’ Example -3: Co-ordinates of the focus of the parabola x2 –4x –8y –4 = 0 are (A) (0, 2) (B) (2, 1) (C) (1, 2) (D) (-2, -1) Solution: (x –2)2 = 8(y + 1) Focus:- X = 0 x –2 = 0 x = 2 Similarly Y = a y + 1 = 2 y = 1 Focus is (2, 1) Hence (B) is the correct answer. Example -4: If the point (2, 3) is the focus and x = 2y + 6 is the directrix of a parabola, find (I) The equation of the axis (ii) The co-ordinates of the vertex (iii) Length of the latus rectum (iv) Equation of the latus rectum Solution: (i) We know that the axis of a parabola is the line through the focus and perpendicular to the directrix. The equation of any line passing through the focus (2, 3) is y – 3 = m(x-2) y – mx = 3 –2m. If the line be perpendicular to the directrix x – 2y = 6 we have, m + 2=0 m = –2 Hence the equation of the axis is y – 3 = –2(x-2) 2x + y = 7 (ii) The co-ordinates of the point of intersection (say) A of the directrix x – 2y = 6 and the axis 2x + y = 7 are obtained by solving the two equations; thus they are (4, -1). Since the vertex O is the middle point of A (4, –1) and the focus S (2, 3); the co-ordinates of the vertex are , 2 1 3 , 2 2 4 i.e. (3, 1). (iii) Since OS = 5 ) 3 1 ( ) 2 3 ( 2 2 , the length of the latus rectum = 4OS = 5 4 .
- 4. (iv) Since the latus rectum is the line through the focus parallel to the directrix, its equation is x-2y+c=0, where c is given by 2 – 2.3 + c = 0, i.e. c = 4 1.4 STANDARD EQUATION OF A PARABOLA: Let S be the focus, ZM the directrix and P the moving point. Draw SZ perpendicular from S on the directrix. Then SZ is the axis of the parabola. Now the middle point of SZ, say A, will lie on the locus of P, i.e., AS = AZ. Take A as the origin, the x-axis along AS, and the y- axis along the perpendicular to AS at A, as in the figure. Let AS = a, so that ZA is also a. Let (x, y) be the coordinates of the moving point P. Then MP = ZN = ZA + AN = a + x. But by definition MP = PS MP2 = PS2 So that, (a + x)2 = (x – a)2 + y2 . Hence, the equation of parabola is y2 = 4ax. Z M S X N P A Y 2.1 DIFFERENT FORMS OF THE PARABOLA Form y2 = 4ax Y2 = –4ax x2 = 4ay x2 = –4ay Shape of the parabola X Y S L L A N P(x,y) (0,0) (a,0) x a 2 y 4ax x x = a A (0, 0) P (x, y) Y 2 y 4ax x y = -b S (0, a) L' L A (0, 0) P (x, y) Y 2 x 4ay x y =a S (0, -a) L' L A(0, 0) P (x, y) Y 2 x 4ay Vertex (0, 0) (0, 0) (0, 0) (0, 0) Focus (a, 0) (–a, 0) (0, a) (0, –a) Equation of Directrix x = –a x = a y = –a y = a Equation of axis y = 0 y = 0 x = 0 x = 0 Tangent at the vertex x = 0 x = 0 y = 0 y = 0 Parametric Equation x = at2 , y = 2at x = –at2 , y = 2at x = 2at, y = at2 x = 2at, y = –at2 Focal Distance of (x1, y1) x1 + a –x1 + a y1 + a –y1 + a Equation of Tangent (a) At the point (x1, y1) yy1 = 2a(x + x1) yy1 = –2a(x + x1) xx1 = 2a(y + y1) xx1 = –2a(y + y1) (b) in terms of m y = mx + a/m y = mx – a/m x = my + a/m x = my – a/m (c) in terms of t yt = x + at2 yt = –x + at2 xt = y + at2 xt = –y + at2 Equation of Normal (a) At the point (x1, y1) y – y1 = a 2 y1 (x – x1) y – y1 = a 2 y1 (x – x1) x–x1 = a 2 x1 (y–y1) x – x1 = a 2 x1 (y – y1) (b) in terms of m y = mx – 2am – am3 y = mx +2am + am3 x= my –2am – am3 x = my +2am + am3 (c) in terms of t y=–tx+2at+at3 y = tx + 2at + at3 x=–ty+ 2at+ at3 x = ty + 2at + at3
- 5. Latus Rectum of the Parabola: Let the given parabola be y2 = 4ax. In the figure LSL is the latus rectum. Also by definition, LSL =2 a 4 a . a 4 = double ordinate through the focus S. y2 = 4ax L S(a, 0) L x A y Notes: Any chord of the parabola y2 = 4ax which is perpendicular to its axis is called the double ordinate. Two parabolas are said to be equal when their latus recta are equal. Illustration 2: Find the equation of parabola whose focus is (1, - 1), and whose vertex is (2, 1). Also find equation of its axis and latus rectum. Solution: As we know that vertex is mid point of focus and point of intersection of directrix with the axis so point P(h, k) is 3 h 2 2 1 h 1 1 3 2 k k P is (3, 3) slope of axis = 2 1 2 ) 1 ( 1 equation of axis = y – 1 = 2 (x – 2) 2x – y = 3 P O S equation of directrix y – 3 = - 2 1 (x – 3) x + 2y = 9 equation of parabola is (x – 1)2 + (y + 1)2 = 5 ) 9 y 2 x ( 2 equation of latus roctum y + 1 = 2 1 (x – 1) x + 2y + 1 = 0 Parametric Equations of a Parabola: If the coordinates of any point (x, y) on a curve can be expressed as functions of a variable t, given by x = (t), y = (t) ……(1) Then the equations in (1) are said to be parametric equations of the curve where ‘t’ is called the parameter. Clearly x = at2 , y = 2at satisfy the equation y2 = 4ax for all real values of t. Hence the parametric equations of the parabola y2 = 4ax are x = at2 , y = 2at, where t is the parameter. Also, (at2 , 2at) is a point on the parabola y2 = 4ax for all real values of t. This point is also described as the point ‘t’ on the parabola. Example -6: The tangents at the points (a 2 1 t , 2at1), (a 2 2 t , 2at2) on the parabola y2 = 4ax are at right angles if (A) t1t2 = -1 (B) t1t2 = 1 (C) t1t2 = 2 (D) t1t2 = -2 Solution: 1 t 1 . t 1 2 1 t1t2 = -1 Hence (A) is the correct answer.
- 6. Example -7: If P(a 2 1 t , 2at1) and Q (a 2 2 t , 2at2) are two variable points on the curve y2 = 4ax and PQ subtends a right angle at the vertex, then t1t2 is equal to (A) –1 (B) –2 (C) –3 (D) –4 Solution: Slope of OP = 1 t 2 Slope of OQ = 2 t 2 Given OP OQ 1 t 2 . 2 t 2 = –1 t1t2 = –4 Hence (D) is the correct answer. Focal Chord: Any chord to the parabola y2 = 4ax which passes through the focus is called a focal chord of the parabola y2 = 4ax. Let y2 = 4ax be the equation of a parabola and (at2 , 2at) a point P on it. Suppose the coordinates of the other extremity Q of the focal chord through P are (at1 2 , 2at1). Then, PS and SQ, where S is the focus (a, 0) have the same slopes. a at 0 at 2 a at 0 at 2 2 1 1 2 tt1 2 – t = t1 t2 - t1 (tt1 + 1)(t1 – t) = 0. Hence t1 = -1/t, i.e. the point Q is (a/t2 , -2a/t). y2 = 4ax Q(at1 2 , 2at1) S(a, 0) P(at2 , 2at) x A y i.e. the extremities of a focal chord of the parabola y2 = 4ax may be taken as the points t and -1/t. Example -8: Prove that the circle with any focal chord of the parabola y2 = 4ax as its diameter always touches its directrix. Solution: Let AB be a focal chord. If A is (at2 , 2at), then B is t a 2 , t a 2 . Equation of the circle with AB as diameter is (x - at2 ) 2 t a x + (y - 2at) t a 2 y = 0 For x = -a, this gives 2 2 2 2 t t 1 a + y2 - 2ay t 1 t - 4a2 = 0 a2 2 t 1 t + y2 - 2ay(t - 1/t) = 0 [y - a(t - 1/t)]2 = 0, which has equal roots. x + a = 0 is a tangent to the circle with diameter AB. Example -9: In the parabola y2 = 4ax, the length of the chord passing through the vertex and inclined to the axis at an angle /4 is (A) 4a 2 (B) 2 a 4 (C) 2a 2 (D) none of these
- 7. Solution: 4 tan at at 2 2 t = 2 P = (4a, 4a) OP = 4 2 a. Hence (A) is the correct answer. X N P(at 2 , 2at) O Y /4 Focal Distance of any Point: The focal distance of any point P (x, y) on the parabola y2 = 4ax is the distance between the point P and the focus S, i.e. PS. Thus the focal distance = PS = PM =ZN = ZA + AN = a + x Z M S X N P A Y Position of a point relative to the Parabola: Consider the parabola : y2 = 4ax. If (x1 , y1) is a given point and y2 1 -4ax1 = 0, then the point lies on the parabola. But when y1 2 - 4ax1 0, we draw the ordinate PM meeting the curve in L. Then P will lie outside the parabola if PM > LM, i.e., PM2 – LM2 > 0 Now, PM2 = y1 2 and LM2 = 4ax1 by virtue of the coordinates of L satisfying the equation of the parabola. Substituting these values in equation of parabola, the condition for P to lie outside the parabola becomes y1 2 - 4ax1 > 0. P(x1, y1) M L X A y Interior Exterior Similarly, the condition for P to lie inside the parabola is y1 2 -4ax1 < 0. Example -10:The coordinates of a point on the parabola y2 = 8x, whose focal distance is 4, are (A) 2 , 2 1 (B) (1, 2 2 ) (C) (2, 4) (D) none of these Solution: Focal distance of a point P (x, y) on y2 = 4ax is (x + a). 4 = x + 2 x = 2 y2 = 8 2 = 16 y = 4 Hence (C) is the correct answer. 1. GENERAL EQUATION OF A PARABOLA Let (h, k) be the focus S and lx + my + n = 0 the equation of the directrix ZM of a parabola. Let (x,y) be the coordinates of any point P on the parabola. Then the relation, PS = distance of P from ZM, gives (x – h)2 + (y – k)2 = (lx + my + n)2 / (l2 + m2) 2 (mx ly) 2gx 2fy d 0 This is the general equation of a parabola. O Z M P(x, y) S(h, k) X Y
- 8. Note: From the general equation of the parabola it is clear that the second-degree terms in the equation of a parabola form a perfect square. The converse is also true, i.e. if in an equation of the second degree, the second-degree terms form a perfect square then the equation represents a parabola, unless it represents two parallel straight lines. Special case: Let the vertex be (, ) and the axis be parallel to the x-axis. Then the equation of parabola is given by (y - )2 = 4a (x – ) which is equivalent to x = Ay2 + By + C Similarly, when the axis is parallel to the y-axis and vertex be(, ), the equation of parabola is given by (x - )2 = 4a (y – ) which is equivalent to y = A’x2 + B’x + C’ Illustration 7: Find the vertex, axis, directrix, tangent at the vertex and the length of the latus rectum of the parabola 2y2 + 3y - 4x - 3 = 0. Solution: The given equation can be re-written as 32 33 x 2 4 3 y 2 which is of the form Y2 = 4aX. Hence the vertex is 4 3 , 32 33 The axis is 0 4 3 y y = -3/4 The directrix is X + a = 0 x + 32 33 + 2 1 = 0 x = 32 49 The tangent at the vertex is 0 32 33 x x = - 32 33 Length of the latus rectum = 4a = 2. INTERSECTION OF A STRAIGHT LINE WITH THE PARABOLA Points of Intersection of a Straight Line with the Parabola: Points of intersection of y2 = 4ax and y = mx + c are given by (mx + c)2 = 4ax i.e. m2 x2 + 2x(mc - 2a) + c2 = 0 ……(1) Since (1) is a quadratic equation, the straight line meets the parabola in two points (real, coincident, or imaginary). The roots of (1) are real or imaginary according as {2(mc – 2a)}2 4m2 c2 is positive or negative, i.e. according as –amc +a2 is positive or negative, i.e. according as mc is less than or greater than a. Note: When m is very small, one of the roots of equation (1) is very large; when m is equal to zero, this root is infinitely large. Hence every straight line parallel to the axis of the parabola meets the curve in one point at a finite distance and in another point at an infinite distance from the X Y ( , ) X Y ( , )
- 9. vertex. It means that a line parallel to the axis of the parabola meets the parabola only in one point. Length of the Chord: As in the preceding article, the abscissae of the points common to the straight line y = mx + c and the parabola y2 = 4ax are given by the equation m2 x2 + ( 2mc – 4a) x + c2 = 0. If (x1, y1) and (x2, y2) are the points of intersection, then (x1 - x2)2 = (x1 + x2)2 - 4x1 x2 = 4 m mc a 16a 2 m 2 4c 4 m 2 2a mc 4 and (y1 - y2) = m(x1 - x2) Hence, the required length = mc a a m 1 m 4 2 2 2 1 2 2 2 1 2 2 1 x x m 1 x x y y TANGENT TO A PARABOLA Tangent at the Point (x1, y1): Let the equation of the parabola be y2 = 4ax. Hence, value of dx dy at P(x1, y1) is 2a y1 and the equation of the tangent at P is y - y1 = 1 y 2a (x - x1) i.e. yy1 = 2a(x - x1) + y1 2 yy1 = 2a(x + x1) Tangent in Terms of m: Suppose that the equation of a tangent to the parabola y2 = 4ax ……(1) is y = mx + c ……(2) The abscissae of the points of intersection of (1) and (2) are given by the equation (mx + c)2 = 4ax. But the condition that the straight line (ii) should touch the parabola is that it should meet the parabola in coincident points (mc - 2a)2 = m2 c2 ……(3) c = a /m. Hence, y = mx + a/m is a tangent to the parabola y2 = 4ax, whatever be the value of m. Equation (mx +c)2 = 4ax now becomes (mx - a/m)2 = 0 x = 2 m a and y2 = 4ax y = m 2a . Thus the point of contact of the tangent y = mx + a/m is m 2a , m a 2 . Tangent at the Point ‘t’: Let the equation of the parabola is y2 = 4ax. The equation of the tangent at (x1, y1) to this parabola is yy1 = 2a(x + x1). If the point (x1, y1) (at2 , 2at) Equation of tangent becomes y.2at = 2a(x + at2 ) yt = x + at2 . Note: The point of intersection of the tangents at ‘t1’ and ‘t2’ to the parabola y2 = 4ax is (at1t2, a(t1 + t2)). Example -11: Two tangents are drawn from the point (-2, -1) to the parabola y2 = 4x. If is the angle between these tangents, then tan equals (A) 3 (B) 1/3
- 10. (C) 2 (D) 1/2 Solution: Here a = 1. Any tangent is y = mx + m 1 . It passes through (-2, -1) 2m2 –m –1 = 0 m = 1, 2 1 tan = 3 2 1 1 2 1 1 Hence (A) is the correct answer. Example -12: If the line 2x + 3y = 1 touches the parabola y2 = 4ax, find the length of the latus rectum. Solution: Equation of any tangent to y2 = 4ax is m a mx y m2 x – my + a = 0 Comparing it with the given tangent 2x + 3y – 1 = 0, we find 1 a 3 m 2 m2 9 2 3 m a , 3 2 m Hence the length of the latus rectum = 4a = 9 8 ignoring the negative sign for length. Example -13: The angle between tangents to the parabola y2 = 4ax at the point where it intersects with the line x – y –a = 0 is (A) 3 (B) 4 (C) 6 (D) 2 Solution: The given chord id a focal chord as it passes through focus (a, 0) and we know that tangents at the extremities of a focal chord meet at right angles on the directrix. Hence (D) is the correct answer. Example -14: On the parabola y2 = 4ax, three points E, F, G are taken so that their ordinates are in G.P. Prove that the tangents at E and G intersect on the abscissa of F. Solution: Let the points E, F, G be (at1 2 , 2at1), (at2 2 , 2at2), (at3 2 , 2at3) respectively. Since the ordinates of these points are in G.P., t2 2 = t1t3. Tangents at E and G are t1y = x + at1 2 and t3y = x + at3 2 . Eliminating y from these equation, we get x = at1t3 = at2 2 , which is the abscissa of F. Equation of the Tangents from an External Point: Let y2 = 4ax be the equation of a parabola and (x1, y1) an external point P. Then, equations of the tangents from (x1, y1) to the given parabola are given by SS1 = T2 , where S = y2 - 4ax, S1 = y1 2 - 4ax1, T = yy1 - 2a(x + x1) Chord of Contact: Equation to the chord of contact of the tangents drawn from a point (x1, y1), to the parabola y2 = 4ax is T= 0, i.e. yy1 - 2a(x+x1) =0. Equation of a Chord with Midpoint (x1, y1):
- 11. The equation of the chord of the parabola y2 = 4ax with mid point (x1, y1) is T= S1 i.e. yy1-2a(x+x1)= y1 2 -4ax1 Or yy1 - 2ax = y1 2 - 2ax1 . Example -15: Find the equation of the chord of the parabola y2 = 12x which is bisected at the point (5, –7). Solution: Here (x1, y1) = (5 –7) and y2 = 12x, a = 3 The equation of the chord is S1 = T or 2 1 y - 4ax1 = yy1 – 2a (x + x1) or (–7)2 – 12.5 = y(–7) – 6(x+5) 6x + 7y + 19 = 0. NORMAL TO THE PARABOLA Normal at the Point (x1, y1 ): The equation of the tangent at the point (x1, y1) is yy1 = 2a(x + x1). Since the slope of tangent = 2a/y1 , slope of normal is -y1/ 2a . Also it passes through (x1, y1). Hence its equation is y - y1 = 1 1 x x 2a y . . . . . (i) Normal in Terms of m: In equation (i), put m 2a y1 so that y1 = -2am and x1 = 2 2 1 am 4a y , then the equation becomes y = mx - 2am - am3 . . . . . (ii) where m is a parameter. Equation (ii) is the normal at the point (am2 , -2am) of the parabola. Notes: If this normal passes through a point (h, k), then k = mh – 2am - am3 . For a given parabola and a given point (h, k) , this cubic in m has three roots say m1, m2, m3 i.e. from (h, k) three normals can be drawn to the parabola whose slopes are m1, m2, m3 . For the cubic, we have m1+ m2 + m3 = 0 m1 m2 +m2 m3 +m3 m1 = (2a-h) /a m1 m2 m3 = - k/a If we have an extra condition about the normals drawn from a point (h, k) to a given parabola y2 =4ax then by eliminating m1, m2, m3 from these four relations between m1, m2, m3, we can get the locus of (h, k). Since the sum of the roots is equal to zero, the sum of the ordinates of the feet of the normals from a given point is zero. Normal at the Point ‘t’: Equation of the normal to y2 = 4ax at the point (x1, y1) is y – y1 = – a 2 y1 (x – x1) If (x1, y1) (at2 , 2at) Equation of normal becomes y – 2at = – a 2 at 2 (x – at2 ) y = –tx + 2at + at3 . Notes: If normal at the point t1 meets the parabola again at the point t2, then t2 = -t1 –2/t1. Point of intersection of the normals to the parabola y2 = 4 ax at (at1 2 , 2at1) and (at2 2 , 2at2) is (2a + a(t1 2 + t2 2 + t1t2),– at1t2(t1+ t2)) .
- 12. Example -16: Find the locus of the foot of the perpendicular drawn from the vertex on a tangent to the parabola y2 = 4ax. Solution: Any tangent of slope m to the parabola y2 = 4ax is y = mx + a/m . . . (1) The perpendicular from the vertex on (1) is x + my = 0 . . . . (2) By eliminating m between (1) and (2) we obtain the required locus as xy2 + x3 + ay2 = 0 Example -17: If the normal at the point (at1 2 , 2at1) meets the parabola y2 = 4ax again at the point (at2 2 , 2at2), prove that t2 = – 1 1 t 2 t Solution: The equation of the normal at (at1 2 , 2at1) is y = –t1x + 2at1 + at1 3 . Since the normal passes through the point (at2 2 , 2at2), we have, 2at2 = -t1. at2 2 +2at1 + at1 3 2a (t1 – t2) = at1 t2 2 – at1 3 = at1 ) t t ( 2 1 2 2 = at1(t2 – t1) (t2 + t1) 2 = –t1 (t2 + t1), as t1 t2 t2 + t1 = 1 t 2 t2 = –t1 1 t 2 Example -18: The normal at any point P to a parabola y2 = 4ax meets its axis at G. Q is another point on the parabola such that QG is perpendicular to the axis of the parabola. Prove that QG2 – PG2 = constant. Solution: Let P be the point (at2 , 2at) Equation of normal at P is y = –tx + 2at + at3 ……(1) Equation (1) meets the x-axis at G whose coordinates are (2a + at2 , 0) PG2 = 4a2 + 4a2 t2 ……(2) Since QG is perpendicular to the axis, the abscissa of Q is the same as that of G and its ordinate is QG. Point Q is (2a + at2 , QG) But Q lies on the parabola y2 = 4ax. y 2 = 4ax G P(at 2 , 2at) (0, 0) Q QG2 = 4a(2a + at2 ) = 8a2 + 4a2 t2 ……(3) QG2 – PG2 = 4a2 = constant by (2) and (3). Subtangent and Subnormal:
- 13. Let the tangent and normal at any point P (x1, y1) on the parabola y2 = 4ax meet the axis in T and G respectively. Then PT is called the length of the tangent at P and PG is called the length of the normal at P. NT is called the subtangent and NG the subnormal at P. The coordinates of T and G can be easily found by putting x = 0 in the equations of the tangent and normal at P. It is evident from figure that Subtangent = NT = twice the abscissa of P Subnormal = NG = 2a = semilatus rectum y 2 = 4ax G P(x1, y1) A Q X T (–x1, 0) S (a, 0) (x1, 0) N (x1 + 2a, 0) X yy1 = 2a(x + x1) Y Y Example -19: If P is a point on the parabola y2 = 4ax such that the subtangent and subnormal at P are equal, find the coordinates of P. Solution: Let P (x, y) be the required point. Length of subtangent = twice the abscisse of P = 2x Length of subnormal = 2a Since subtangent = subnormal, 2x = 2a x = a y2 = 4a2 y = 2a The required points are (a, 2a) and (a, –2a). PROPERTIES OF THE PARABOLA P y 2 = 4ax S (Focus) x A T M y K Z M P N G (i) The tangent at any point P on a parabola bisects the angle between the focal chord through P and the perpendicular from P on the directrix. In the given figure MPT = TPS : Similarly the normal at any point on a parabola bisects the angle between the focal chord and the line parallel to the axis through that point. (ii) The portion of a tangent to a parabola cut off between the directrix and the curve subtends a right angle at the focus. In the given figure SP is perpendicular to SK i.e. KSP = 900 . (iii) Tangents at the extremities of any focal chord intersect at right angles on the directrix. (iv) Any tangent to a parabola and the perpendicular on it from the focus meet on the tangent at the vertex.
- 14. Pole and Polar To find the equation of the polar of the point (x1, y1) with respect to the parabola y2 = 4ax. Let Q and R be the points in which any chord drawn through the point P, whose coordinates are (x1, y1), meets the parabola. Let the tangents at Q and R meet in the point whose coordiantes are (h, k). R’ T’ P Q R T (h, k) (x1, y1) R’ T’ P Q’ R T (h, k) (x1, y1) Q We require the locus of (h, k). Since QR is the chord of contact of tangents from (h, k) its equation is ky = 2a (x + h) Since this straight line passes through the point (x1, y1) we have ky1 = 2a (x 1+ h) ….(i) Since the relation (i) is true, it follows that the point (h, k) always lies on the straight line yy1 = 2a(x + x1) ….(ii) Hence (ii) is the equation to the polar of (x1, y1). Example -20:Prove that the locus of poles of focal chord of the parabola y2 = 4ax is the directrix. Solution: Let (h, k) be the pole. Then the equation of the chord is ky = 2a(x + h). Since it is a focal chord, it passes through the focus (a, 0). 2a(a + h) = 0 or a + h = 0 Hence locus of pole (h, k) is x + a = 0, which is the directrix.
- 15. ASSIGNMENT 1. The focus of the parabola x2 – 8x + 2y + 7 = 0 is (A) 1 0, - 2 æ ö ç ÷ è ø (B) 9 4, 2 æ ö ç ÷ è ø (C) (4, 4) (D) 9 -4, - 2 æ ö ç ÷ è ø Solution: Given parabola is (x – 4)2 = –2 2 9 y , focus is given by x – 4 = 0, y – 2 9 = 2 1 focus is (4, 4). Hence (C) is the correct answer. 2. For the parabola y2 = 4ax, the ratio of the subtangent to the abscissa is (A) 1 : 1 (B) 2 : 1 (C) x : y (D) x2 : y2 Solution: x y y x gent tan Sub & 2yy = 4a y = y a 2 from y2 = 4ax ax 2 ax 4 x gent tan Sub = 1 2 . Hence (B) is the correct answer. 3. The length of subnormal to the parabola y2 = 4ax at any point is equal to (A) a 2 (B) 2 2 a (C) a/ 2 (D) 2a Solution: Length of subnormal at point (x1, y1) = y1 a 2 y a 2 . y dx dy 1 1 . Hence (D) is the correct answer. 4. The length of the latus rectum of the parabola x = ay2 + by + c is (A) 4 a (B) 3 a (C) a 1 (D) a 4 1 Solution: x = a c a 4 b a 4 b y a b y 2 2 2 2 or, a c a 4 b x a 2 b y 2 2 or, a 4 ac 4 b x a 1 a 2 b y 2 2 the latus rectum = a 1 Hence (C) is the correct answer. 5. The number of normals drawn from a point (3, 0) to the parabola y2 = 4x is /are (A) 1 (B) 2 (C) 3 (D) none of these Solution: y = mx – 2m – m3, this passes through (3, 0) hence
- 16. 0 = 3m – 2m – m3 m3 – m = 0 m = 0, m = 1 Hence three normals can be drawn. Hence (C) is the correct answer. 6. If a normal chord to the parabola y2 = 4x is drawn at (1, 2), then the chord meets the parabola again at: (A) (9, 6) (B) (9, –6) (B) (6, 6) (D) none of these Solution: For point (1, 2) value of t = 1, the value of t2 for other end of the normal is t2 = –1 – 1 2 = –3. So point is (at2, 2at) = (9, –6) Hence (B) is the correct answer. 7. The area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum is (A) 12 sq. units (B) 16 sq. units (C) 18 sq. units (D) 24 sq. units Solution: Required area is given by = 2 1 (12x3) = 18 sq. units Hence (C) is the correct answer. 8. If the line x+ y –1 = 0 touches the parabola y2 = kx , then the value of k is (A) 4 (B) –4 (C) 2 (D) –2 Solution: Any tangent to y2 = kx is ; y = mx +k/4m Comparing it with the given line y = 1 – x, we get, m = -1 and k/4m = 1 k = -4 Alternative Solution: If x+ y –1 = 0 touches y2 = kx, then y2 = k(1-y) would have equal roots k2 + 4k = 0 k = 0 or -4 but k 0, hence k = -4 Hence (B) is the correct answer. 9. The coordinates of the point on the parabola y = x2 + 7x +2, which is nearest to the straight line y = 3x – 3 are (A) ( -2, -8) (B) ( 1, 10) (C) ( 2, 20) (D) ( -1, -4) Solution: Any point on the parabola is (x , x2+ 7x + 2) Its distance from the line y = 3x –3 is given by P = 10 5 x 4 x 1 9 3 2 x 7 x x 3 2 2 = 10 5 x 4 x2 (as x2 +4x + 5 > 0 for all x R) dx dP = 0 x = -2 The required point ( -2, -8) Hence (A) is the correct answer.
- 17. 10. The point (1, 2) is one extremity of focal chord of parabola y2 = 4x. The length of this focal chord is (A) 2 (B) 4 (C) 6 (D) none of these Solution: The parabola y2 = 4x. Here a =1 and focus is (1, 0). The focal chord is ASB. This is clearly latus rectum of parabola, its value = 4. Hence (B) is the correct answer. S(1, 0) A(1, 2) X y B 11. If AFB is a focal chord of the parabola y2 = 4ax and AF = 4, FB = 5, then the latus- rectum of the parabola is equal to (A) 9 80 (B) 80 9 (C) 9 (D) 80 Solution: FA = 4, FB = 5 We know that FB 1 AF 1 a 1 a = 9 20 4a = 9 80 . Hence (A) is the correct answer. A B F (a, 0) O 12. Vertex of the parabola whose parametric equation is x = 2 t - t + 1, y = 2 t + t + 1; t R, is (A) (1, 1) (B) (2, 2) (C) (1/2, 1/2) (D) (3, 3) Ans. (A) Solution. x = 2 t - t + 1, y = 2 t + t + 1 x + y = 2( 2 t + 1) and y – x = 2t x y 2 = 1 + 2 y x 2 2 (y x) = 2(x + y) – 4 2 (y x) = 2 (x + y – 2). Vertex will be the point where lines y – x = 0 and x + y – 2 = 0 meet, i.e., the point (1,1). 13. If the vertex of the parabola y = x2 – 8x + c lies on x-axis, then the value of c is (A) –16 (B) –4 (C) 16 (D) 4 Solution: The parabola is y = (x – 4)2 + c – 16. So the vertex is (4, c – 16). As vertex is on x-axis, c = 16. 14. If (2, 0) is the vertex and y-axis the directrix of a parabola, then its focus is (A) (2, 0) (B) (-2, 0) (C) (4, 0) (D) (-4, 0) Solution: a = distance of vertex from focus = distance of vertex from directrix = 2. So, focus is at (4, 0).
- 18. 15. If the straight line y = mx + c (m > 0) touches the parabola y2 = 8(x + 2), then the minimum value taken by c is (A) 12 (B) 8 (C) 4 (D) 6 Solution: The tangent of slope m must be of the form y = m(x + 2) + m a . So, 2m + m 2 = c c = 2 m 1 m 2 2. So cmin = 4. 16. The length of the latus rectum of the parabola whose focus is (3, 3) and directrix is 3x – 4y – 2 = 0 is (A) 2 (B) 3 (C) 1 (D) 4 Solution: Semilatus rectum = 5 | 2 3 4 3 3 | = 1, so latus rectum = 2. 17. The length of the subtangent to the parabola y2 = 16x at the point whose abscissa is 4, is (A) 2 (B) 4 (C) 8 (D) none of these Solution: The length of subtangent = 2x1 = 2 4 = 8. 18. If the line x + y = 1 touches the parabola y2 –y + x = 0, then the co-ordinates of the point of contact are (A) (1, 1) (B) 2 1 , 2 1 (C) (0, 1) (D) (1, 0) Solution: The tangent at (x1, y1) is 2yy1 – y – y1 + x + x1 = 0 comparing it with x + y – 1 = 0, we get (x1, y1) (0, 1). 19. If the chord y = mx + c subtends a right angle at the vertex of the parabola y2 = 4ax, then the value of c is (A) – 4am (B) 4am (C) –2am (D) 2am Solution: y2 – 4ax c mx y = 0 Coefficient of x2 + Coefficient of y2 = 0 c = –4am. 20. Minimum distance between the curve 2 y = 4x and 2 x + 2 y -12x + 31 = 0 is equal to (A) 21 (B) 26 5 (C) 21 5 (D) 28 5 Ans. (C) Solution. Shortest distance will take place along the common normal.
- 19. (6,0) y x A O C Equation of normal for 2 y = 4x at (a 2 t , 2t) is y = -tx + t + 3 t If it passes through (6, 0), then 3 t - 5t = 0 t = 0, t = 5 . A (5, 2 5 ), C (5, -2 5 ). Now, PA = PC = 1 20 = 21, OP = 6 Minimum distance = ( 21 5 )units. 21. If (x1, y1) and (x2, y2) are the ends of a focal chord of the parabola y2 = 4ax, then x1x2+ y1y2 = (A) –3a2 (B) 3a2 (C) -4a2 (D) 4a2 Solution: x1x2 + y1y2 = a2 2 2 2 1t t + 4a2t1t2 = a2 – 4a2 = –3a2. 22. If y = 2x + 3 is a tangent to the parabola y2 = 24x, then its distance from the parallel normal is (A) 5 5 (B) 10 5 (C) 15 5 (D) none of these Solution: distance = 4 1 48 24 3 m 1 am am 2 m a 2 3 = 15 5 . 23. If the segment intercepted by the parabola y2 = 4ax with the line lx + my + n = 0 subtends a right angle at the vertex, then (A) 4am + n = 0 (B) 4am + 4al + n = 0 (C) 4al + n = 0 (D) al + n = 0 Solution: y2 + 4ax n my lx = 0, now coefficient of x2 + coefficient of y2 = 0 4al + n = 0. 24. The length of a focal chord of the parabola y2 = 4ax making an angle with the axis of the parabola is (A) 4a cosec2 (B) 4a sec2 (C) a cosec2 (D) none of these Solution: length = a|t1 – t2| 4 ) t t ( 2 2 1 Now, t1t2 = –1 and 2 1 t t 2 = tan
- 20. So, t1 + t2 = 2 cot , |t1 – t2| = ) 1 ( 4 ) cot 2 ( 2 = 2 |cosec | So, required length = a 2|cosec | 2|cosec | = 4a cosec2 . 25. Equation of parabola having it’s vertex at A(1,0) and focus at S(3, 0) is (A) 2 y = 4 (x - 1) (B) 2 y = 4 (x + 1) (C) 2 y = 8 (x - 1) (D) 2 y = 8 (x + 1) Ans. (C) Solution. Clearly, axis of parabola is the x-axis. Corresponding value of a = 3 – 1 = 2. Thus, equation of parabola is 2 y = 8(x – 1). 26. Equation of parabola having it’s focus at S(2, 0) and one extremity of it’s latus rectum as (2, 2) is (A) 2 y = 4 (3 - x) (B) 2 y = 4 (1 - x) (C) 2 y = 8 (3 - x) (D) 2 y = 8 (3 - x) Ans. (A) Solution. Clearly, the other extremity of latus rectum is (2, 2). It’s axis is x-axis. Corresponding value of a = 2 0 2 = 1 Hence it’s vertex is (1, 0) or (3, 0) Thus it’s equation is 2 y = 4(x – 1) or 2 y = 4(x – 3). 27. Equation of parabola having the extremities of it’s latus rectum as (3, 4) and (4, 3) is (A) 2 2 2 7 7 x y 6 x y 2 2 2 (B) 2 2 2 7 7 x y 8 x y 2 2 2 (C) 2 2 2 7 7 x y 4 x y 2 2 2 (D) None of these Ans. (B,A) Solution. Focus is 7 7 , 2 2 and it’s axis is the line y = x. Corresponding value of ‘a’ = 1 2 ( 1 1) 4 4 . Let the equation of it’s directrix be y + x + = 0. | 3 4 | 2 2. 4 2 = -6, -8 Thus equation of parabola is 2 2 2 7 7 (x y 6) x y 2 2 2 or 2 2 2 7 7 (x y 8) x x 2 2 2 28. AB is focal chord of 2 x - 2x + y – 2 = 0 whose focus is ‘S’. If AS = 1 then BS is equal to
- 21. (A) 1 1 4 4 1 (B) 1 1 4 1 (C) 1 1 2 4 1 (D) None of these Ans. (B) Solution. 2 x – 2x + y – 2 = 0 2 x - 2x + 1 = 3 - y 2 (x 1) = -(y – 3) Length of it’s latus rectum is 1 unit. Since AS, 1 2 , BS are in H.P., therefore 1 2 = 2.AS.BS AS BS BS = 1 1 I (4 1) 29. Locus of mid-point of chords of the parabola 2 y = 4ax that pass through the point (3a, a) is (A) 2 y + 2ax + ay – 6 2 a = 0 (B) 2 y + 2ax - ay + 6 2 a = 0 (C) 2 y - 2ax + ay + 6 2 a = 0 (D) 2 y - 2ax - ay + 6 2 a = 0 Ans. (D) Solution. Let the mid point of chord be P(h, k) then it’s equation is T = 1 S i.e., yk – 2a(x + h) = 2 k - 4ah It must pass through (3a, a), hence ak – 2a(3a + h) = 2 k - 4ah Thus locus of P is 2 y - 2ax – ay + 6 2 a = 0 30. Locus of trisection point of any double ordinate of 2 y = 4ax is (A) 3 2 y = 4ax (B) 2 y = 6ax (C) 9 2 y = 4ax (D) None of these Ans. (C) Solution. Let AB be a double ordinate, where A (a 2 t , 2at), B (a 2 t , - 2at). If P(h, k) be it’s trisection point, then 3h = 2a 2 t + a 2 t , 3k = 4at – 2at 2 t = h a , t = 3k 2a Thus locus is, 2 2 9k h 4a a i.e., 9 2 y = 4ax. 31. If the line joining the points 2 1 1 A(at ,at ) and 2 2 2 B(at ,at ) passes through C(0, b), then (A) 1 2 1 2 b(t t ) 2at t (B) 1 2 1 2 2b(t t ) at t
- 22. (C) 1 2 1 2 b(t t ) at t (D) None of these Ans. (A) Solution. Equation of line joining A, B is y( 1 2 t t ) = 2(x + a 1 2 t t ) It will pass through (0, b), if b( 1 2 t t ) = 2a 1 2 t t 1 2 t t = 2a b 1 2 t t . 32. If the lines (y – b) = 1 m (x + a) and (y – b) = 2 m (x + a) are the tangents of 2 y = 4ax, then (A) 1 2 m m 0 (B) 1 2 m m 1 (C) 1 2 m m 1 (D) 1 2 m m 1 Ans. (C) Solution. Clearly, both the lines pass through (-a, b) which is a point lying on the directrix of the parabola. Thus, 1 2 m m = -1. Because tangents drawn from any point on the directrix are always mutually perpendicular. 33. Tangents drawn to parabola 2 y = 4ax at the points A and B intersect at C. Ordinate of A, C and B forms (A) a A.P. (B) a G.P. (C) a H.P. (D) None of these Ans. (A) Solution. If A (a 2 1 t , 2a 1 t ), B (a 2 2 t , 2a 2 t ) C (a 1 2 t t , a( 1 2 t t )) Clearly the ordinates of A,C, B are in A.P. 34. Tangents drawn to parabola 2 y = 4ax at the point A and B intersect at C. If ‘S’ be the focus of the parabola then, SA, SC and SB forms (A) a A.P. (B) a G.P. (C) a H.P. (D) None of these Ans. (B) Solution. If A (a 2 1 t , 2a 1 t ), B (a 2 2 t , 2a 2 t ), C (a 1 2 t t , a( 1 2 t t )) Now, SA = a + a 2 1 t , SB = a + a 2 2 t , SC = 2 2 2 1 2 1 2 (at t a) a (t t ) SC = a 2 2 1 2 1 2 (t t 1) (t t ) = a 2 2 2 2 1 2 1 2 t t 1 t t = a 2 2 1 2 (1 t )(1 t ) Clearly, 2 SC = SA.SB 35. Locus of trisection point of any arbitrary double ordinate or the parabola 2 x 4by is
- 23. (A) 2 9x by (B) 2 3x 2by (C) 2 9x 4by (D) 2 9x 2by Ans. (C) Solution. Let A (2bt, b 2 t ), B (-2bt, b 2 t ) be the extremities on the double ordinate AB. If C(h, k) be it’s trisection point, then 3h = 4bt – 2bt, 3k = 2b 2 t + b 2 t . t = 3h 2b , 2 t = k b k b = 2 2 9h 4b Thus locus of C is 9 2 x = 4by 36. Equation of common tangent of 2 y = 4ax and 2 x 4by is (A) x + y – a = 0 (B) x - y – a = 0 (C) x - y + a = 0 (D) x + y + a = 0 Ans. (D) Solution. Let y = mx + a m be a tangent of 2 y = 4ax. It will touch 2 x = 4ay, provided 2 x = 4a a mx m has equal roots. 16 2 2 a m = -16 2 a m m = -1. Thus common tangent is y + x + a = 0. 37. If the line ax + by + c = 0 is a tangent to the parabola 2 y - 4y – 8x + 32 = 0, then (A) 2 4b = a(7a + 2c + 4b) (B) 2 4b = a(7a + c - 4b) (C) 2 4b = a(7a + 2c - b) (D) 2 4b = a(7a + c - b) Ans. (A) Solution. Line will touch the parabola provided 2 y – 4y + 32 = 8( by c) a has equal roots. 4 2 b = 7 2 a + 2a(c + 2b). 38. The straight line y = m(x –a) will meet the parabola 2 y = 4ax in two distinct real points if (A) m R (B) m [ 1,1] (C) m [ ,1] [1 , ) (D) m R ~ {0} Ans. (D) Solution. y = m(x – a) passes through the focus (a, 0) of the parabola. Thus for this to be focal chord m R ~ {0}. 39. Length of the shortest normal chord of the parabola 2 y = 4ax is (A) a 27 (B) 3a 3 (C) 2a 27 (D) None of these Ans. (C) Solution. Let AB be a normal chord where A (a 2 1 t , 2a 1 t ), B (a 2 2 t , 2a 2 t ).
- 24. We have, 2 1 1 2 t t t . 2 2 2 2 2 2 2 1 2 1 2 AB [a (t t )] 4a (t t ) = 2 2 2 1 2 1 2 a (t t ) [(t t ) 4] = 2 2 1 1 2 1 1 2 4 a t t 4 t t = 2 2 3 1 4 1 16a (1 t ) t 2 1 d(AB ) dt = 16 4 2 2 2 3 3 2 1 1 1 1 1 8 1 t [3(1 t ) 2t ] (1 t ) .4t a t = 2 2 2 2 1 1 5 1 a .32(1 t ) (t 2) t 1 t = 2 is indeed the point of minima of A 2 B . Thus, 3 / 2 mini 4a AB (1 2) 2 = 2a 27 units. 40. Mutually perpendicular tangents TA and TB are drawn to 2 y = 4ax, minimum length of AB is equal to (A) 4a (B) 6a (C) 8a (D) 2a Ans. (A) Solution. Chord of contact of mutually perpendicular tangents is always a focal chord. Thus minimum length of AB is 4a. 41. If three distinct and real normals can be drawn to 2 y = 8x from the point (a, 0), then (A) a > 2 (B) a > 4 (C) a (2,4) (D) None of these Ans. (B) Solution. Equation of normal in terms of m is y = mx – 4m – 2 3 m If it passes through (a, 0), then am – 4m - 2 3 m = 0 m(a - 4 - 2 2 m ) = 0 m = 0, 2 m = a 4 2 . For three distinct normals, a – 4 > 0 a > 4. 42. PA and PB are tangents drawn to 2 y = 4ax from arbitrary point P. If the angle between tangents is / 4 , then locus of point P is (A) 2 2 2 y x a 6ax (B) 2 2 2 y x a 6ax (C) 2 2 2 y x a 6ax (D) 2 2 2 y x a 6ax
- 25. Ans. (A) Solution. Let P (h, k) Tangent in terms of m is y = mx + a m 2 m x – my + a = 0. If it passes through (h, k), then 2 m h – mk + a = 0. Since, angle between these tangents is 4 , tan 4 = 1 2 1 2 | m m | |1 m m | 2 2 1 2 1 2 (m m ) (1 m m ) 2 2 1 2 1 2 1 2 (m m ) 4m m (1 m m ) 2 2 2 k 4a a 1 h h h Thus required locus is, 2 y = 4ax + (a + x)2. 43. The parabola 2 y = 4x and the circle 2 2 2 (x 6) y r will have no common tangent if ‘r’ is equal to (A) r 20 (B) r 20 (C) r 18 (D) r ( 20, 28) Ans. (A) Solution. Any normal of parabola is y = -tx + 2t + 3 t If it pass through (6, 0), then -6t + 2t + 3 t = 0 t = 0, 2 t = 4, A (4, 4). y x A O (6,0) Thus for no common tangent AC > 4 16 > r r > 20 44. Parabola 2 y = 4x and the circle having it’s centre at (6, 5) intersect at right angle. Possible point of intersection of these curves can be (A) (9, 6) (B) (2, 8) (C) (1, 2) (D) (3,2 3)
- 26. Ans. (A) Solution. Let the possible point be ( 2 t , 2t) Equation of tangent at this point is yt = x + 2 t It must pass through (6, 5) 2 t - 5t + 6 = 0 t = 2, 3 Thus possible points are (4, 4), (9, 6). 45. Normal PO, PA and PB (‘O’ being the origin) are drawn to 2 y = 4x from P(h, 0). If AOB /2 , then area of quadrilateral OAPB is equal to (A) 12 sq. units (B) 24 sq. units (C) 6 sq. units (D) 18 sq. units Ans. (B) Solution. Let A ( 2 t , 2t), B ( 2 t , -2t) OA OB 2 2 m , m t t 2 t 4 t = 2 y x A O P B Equation of normal AP is y = -2x + 4 + 8 P (6, 0) Thus area of quadrilateral OAPB = 1 2 (OP)(AB) = 1 2 .6.8 = 24 sq. units. 46. Locus of the midpoint of any focal chord of 2 y = 4ax is (A) 2 y = a(x – 2a) (B) 2 y = 2a(x – 2a) (C) 2 y = 2a(x – a) (D) None of these Ans. (C) Solution. Let the midpoint be P(h, k) Equation of this chord is T = 1 S . i.e., yk – 2a(x + h) = 2 k - 4ah It must pass through (a, 0) 2a(a + h) = 2 k - 4ah Thus required locus is 2 y = 2ax – 2 2 a
- 27. 47. Locus of the midpoint of any normal chord of 2 y = 4ax is (A) 2 2 2 2 4a y x a 2 y 2a (B) 2 2 2 2 4a y x a 2 y 2a (C) 2 2 2 2 4a y x a 2 y 2a (D) 2 2 2 2 4a y x a 2 y 2a Ans. (B) Solution. Let AB be a normal chord where A (a 2 1 t ,2 1 t ), B (a 2 2 t ,2 2 t ), If it’s midpoint is P(h, k), then 2h = a( 2 1 t + 2 2 t ) = a[ 2 1 2 (t t ) - 2 1 2 t t ] and 2k = 2a 1 2 (t t ) We also have 2 t = - 1 t - 1 2 t 1 2 t t = 1 2 t and 1 2 t t = - 2 1 t - 2. 1 t = - 2a k and h = a 2 1 2 1 2 t 2 t Thus required locus is x = a 2 2 2 2 4a y 2 y 2a . 48. The locus of the centre of a circle that passes through P(a, b) and touch the line y = mx + c (it is given that b ma c ) is (A) a straight Circle (B) circle (C) parabola (D) hyperbola Ans. (C) Solution. Let O(h, k) be the centre of circle Clearly distance of O from P and the line y = mx + c will be equal Thus locus of P will be a parabola. 49. Slope of the normal chord of 2 y = 8x that gets bisected at (8, 2) is (A) 1 (B) -1 (C) 2 (D) -2 Ans. (C) Solution. Let AB be the normal chord where A (2 2 1 t , 4 1 t ), B (2 2 2 t , 4 2 t ), It’s slope = 1 2 2 t t We also have 2 1 1 2 t t t and 16 = 2( 2 2 1 2 t t ), 4 = 4 1 2 (t t ) 1 2 t t = 1
- 28. Thus slope is 2. 50. Radius of the circle that passes through origin and touches the parabola 2 y = 4ax at the point (a, 2a) is (A) 5 a 2 (B) 2 2a (C) 5 2a (D) 3 2a Ans. (A) Solution. Equation of tangent of parabola at (a, 2a) is y.2a = 2a(x + a), i.e., y – x – a = 0. Equation of circle touching the parabola at (a, 2a) is 2 2 (x a) (y 2a) (y – x – a) = 0 Since, it passes through (0, 0), therefore 2 2 a 4a (-a) = 0 = 5a. Thus required circle is 2 2 x y - 7ax + ay = 0 It’s radius = 2 2 49 a 5 a a. 4 4 2 51. The length of latus rectum of the parabola, whose focus is (2 cos2 , a cos2 ) and directrix is the line y = a, is (A) |4a 2 cos | (B) |4a 2 sin | (C) |4a cos2 | (D) |a cos2 | Ans. (B) Solution. Distance of focus from directrix is |a cos2 – a| Thus length of latus rectum is |4a 2 sin |. 52. If the line y = mx + c is a tangent to 2 y = 4mx, then distance of this tangent from the parallel normal is (A) 2 1 m (B) 2 2 m (C) 2 3 / 2 (1 m ) (D) 2 3 / 2 (2 m ) Ans. (C) Solution. Equation of parallel normal is y = mx - 2 2 4 m m Required distance = 2 4 2 | c 2m m | 1 m Here, c = 1 Thus required distance = (1 + 2 m )3/2 53. Angle between the tangents drawn to 2 y = 4x at the points where it is intersected by the line y = x – 1 is equal to (A) 4 (B) 3
- 29. (C) 6 (D) 2 Ans. (D) Solution. The line y = x – 1 passes through (1, 0). that means it is a focal chord. Hence the required angle is /2. 54. Tangents PA and PB are drawn to parabola 2 y = 4ax. If slope of bisector of the angle PAB is 3 , then locus of point ‘p’ is (A) 2 2 2 y x 3a 2ax (B) 2 2 2 y 3x 3a 2ax (C) 2 2 2 y x 3a 10ax (D) 2 2 2 y 3x 3a 10ax Ans. (D) Solution. Slope of tangents will be tan 3 , tan 3 , Let A (a 2 1 t , 2a 1 t ), B (a 2 2 t , 2a 2 t ), P 1 2 1 2 [at t ,a(t t )] We have 1 t = cot 3 , 2 t = cot 3 , 1 2 1 1 .cot 1 .cot 1 3 3 t ,t 1 1 cot cot 3 3 1 2 1 2 3 t t 3 cot 1 3t 1 3t 2 1 1 2 2 1 2 1 3 3t t 3t t t 3 3t t 3t 1 2 1 2 2 3 2 3t t 2(t t ) 2 2 2 2 2 1 2 1 2 1 2 3(1 t t ) (t t ) (t t ) 4t t Thus locus of P is 2 2 2 y 3x 3a 10ax 55. Parabolas 2 1 y 4a(x c ) and 2 2 x 4a(y c ) where 1 c and 2 c are variable, are such that they touch each other. Locus of their point of contact is (A) xy = 2 2 a (B) xy = 4 2 a (C) xy = 2 a (D) None of these Ans. (B) Solution. Let P(x, y) be the point of contact. At ‘P’ both of them must have same slope. We have, 2y dy dx = 4a, 2x = 4a dy dx . Eliminating dy dx , we get xy = 4 2 a 56. The circle 2 2 x y + 2gx + m 2fy + c = 0, cuts the parabola 2 x = 4ay at points i i i A (x ,y ), i = 1, 2, 3, 4; then
- 30. (A) i y = 0 (B) i y = -4(f + 2a) (C) i x = - 4(g + 2a) (D) i x = - 2(g + 2a) Ans. (B) Solution. Putting y = 2 x 4a in the equation of circle, we get 4 2 2 2 x fx x 2gx c 0 16a 2a 4 2 2 2 2 x x (16a 8af) 32a gx 16a c 0 1 2 x x = 8a(f + 2a), 2 2 1 2 3 1 2 3 4 x x x 32a g, x x x x 16a c 2 2 i 1 1 1 2 1 1 y x [( x ) 2 x x ] 4a 4a = -4(f + 2a) 57. Maximum number of common normals of 2 y = 4ax and 2 x = 4by can be equal to (A) 3 (B) 4 (C) 6 (D) 5 Ans. (D) Solution. Normals to 2 y = 4ax and 2 x = 4by in terms of ’m’ are y = mx – 2am – a 3 m and y = mx + 2b + 3 b m . For a common normal 2b + 2 b m - 2am + a 3 m = 0 a 5 m + 2a 3 m + 2b 2 m + b = 0 That means there can be atmost 5 common normals. 58. If two different tangents of 2 y = 4x are the normals to 2 x =4by, then (Ref.Parabola P.K.Sharma Pg.C3.8 Q.18) (A) 1 | b | 2 2 (B) 1 | b | 2 2 (C) 1 | b | 2 (D) 1 | b | 2 Ans. (B) Solution. Tangent to 2 y = 4x in terms of m is y = mx + 1 m . Normal to 2 x = 4by in terms of m is y = mx + 2b + 2 b m . If these are same lines, then 1 m = 2b + 2 b m 2b 2 m – m + b = 0 For two different tangents, we must have 1 – 8 2 b > 0
- 31. |b| < 1 8 59. Length of the latus rectum of the parabola whose parametric equation is x = 2 t + t + 1, y = 2 t - t + 1, where t R , is equal to (Ref.Parabola P.K.Sharma Pg.C3.9 Q.21) (A) 8 (B) 4 (C) 2 (D) None of these Ans. (D) Solution. x = 2 t + t + 1, y = 2 t - t + 1 x + y = 2( 2 t + 1) and x – y = 2t (x y) 2 1 + 2 x y 2 2 (x y) 2 (x + y) – 4 = 2(x + y – 2) Comparing it with 2 y = 4ax, we get length of latus rectum = 2 units 60. Minimum distance between the curves 2 y = x – 1 and 2 x = y – 1 is equal to (A) 3 2 4 (B) 5 2 4 (C) 7 2 4 (D) 1 2 4 Ans. (A) Solution. Both curves are symmetrical about the line y = x. If line AB is the line of shortest distance then at A and B slopes of curves should be equal to one. For 2 y = x – 1, dy 1 dx 2y = 1 y = 1 2 , x = 5 4 B 1 5 , 2 4 . B A O x y y=x Hence minimum distance AB, = 2 2 5 1 5 1 3 2 4 2 4 2 4 units.