# (4) Parabola theory. Module-3pdf

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### (4) Parabola theory. Module-3pdf

• 1. Total No. of questions in Parabola are- InchapterExamples.......................................38 SolvedExamples...........................................24 TotalNo. ofquestions............................. ....62 PARABOLA
• 2. 1. DEFINITION A parabola is the locus of a point which moves in such a way that its distance from a fixed point is equal to its perpendicular distance from a fixed straight line. 1.1 Focus : The fixed point is called the focus of the Parabola. 1.2 Directrix : The fixed line is called the directrix of the Parabola. 2. TERMS RELATED TO PARABOLA 2.1 Eccentricity : If P be a point on the parabola and PM and PS are the distances from the directrix and focus S respectively then the ratio PS/PM is called the eccentricity of the Parabola which is denoted by e. Note: By the definition for the parabola e = 1. If e > 1  Hyperbola, e = 0  circle, e < 1  ellipse 2.2 Axis : A straight line passes through the focus and perpendicular to the directrix is called the axis of parabola. (focus) 2.3 Vertex : The point of intersection of a parabola and its axis is called the vertex of the Parabola. NOTE: The vertex is the middle point of the focus and the point of intersection of axis and directrix 2.4 Focal Length (Focal distance) : The distance of any point P (x, y) on the parabola from the focus is called the focal length. i.e. The focal distance of P = the perpendicular distance of the point P from the directrix. 2.5 Double ordinate : The chord which is perpendicular to the axis of Parabola or parallel to Directrix is called double ordinate of the Parabola. 2.6 Focal chord : Any chord of the parabola passing through the focus is called Focal chord. 2.7 Latus Rectum : If a double ordinate passes through the focus of parabola then it is called as latus rectum. 2.7.1Length of latus rectum : The length of the latus rectum = 2 x perpendicular distance of focus from the directrix. 3. STANDARD FORM OF EQUATION OF PARABOLA If we take vertex as the origin, axis as x- axis and distance between vertex and focus as 'a' then equation of the parabola in the simplest form will be- y2 = 4ax y M L(a,2a) L’ S (a,0) x A Z Directrix x = a Focus axis Focal chord Double ordinate Latus Rectum Vertex F o c a l d is ta n c e y’ x+ a = 0 P (a,-2a) L’ L
• 3. 3.1 Parameters of the Parabola y2 = 4ax (i) Vertex A  (0, 0) (ii) Focus S  (a, 0) (iii) Directrix  x + a = 0 (iv) Axis  y = 0 or x– axis (v) Equation of Latus Rectum  x = a (vi) Length of L.R.  4a (vii) Ends of L.R.  (a, 2a), (a, – 2a) (viii) The focal distance  sum of abscissa of the point and distance between vertex and L.R. (ix) If length of any double ordinate of parabola y2 = 4ax is 2  then coordinates of end points of this Double ordinate are           , 4a 2 and           , 4a 2 . 3.2 Other standard Parabola : Examples based on Standard form of an equation of Parabola Ex.1 If focus of a parabola is (3,–4) and directrix is x + y – 2 = 0, then its vertex is (A) (4/15, – 4/13) (B) (–13/4, –15/4) (C) (15/2, – 13/2) (D) (15/4, – 13/4) Q1 P1 Y S X M Z S P1 Q1 X M A Z y2 = – 4 ax x2 = 4ay x2 = – 4ay Equationof Vertex Axis Focus Directrix Equation of Lengthof Parabola Latus rectum Latus rectum y2 = 4ax (0,0) y= 0 (a,0) x = –a x= a 4a y2 = – 4ax (0,0) y= 0 (–a, 0) x = a x= –a 4a x2 = 4ay (0,0) x=0 (0,a) y = a y = a 4a x2 = – 4ay (0,0) x=0 (0, –a) y = a y = –a 4a
• 4. Sol. First we find the equation of axis of parabola, which is perpendicular to directrix. So its equation is x – y + k = 0. It passes through focus S (3, –4)  3 – (–4) + k = 0  k = –7 Let Z is the point of intersection of axis and directrix. Solving equation x + y –2 = 0 and x –y–7=0 gives Z (9/2, –5/2) Vertex A is the mid point of Z and S  A                2 2 5 4 , 2 9 3 2 = A A        4 13 , 4 15 Ans.[D] Ex.2 The equation of the parabola whose vertex is (–3, 0) and directrix is x + 5 = 0 is- (A) y2 = 8(x + 3) (B) y2 = – 8(x + 3) (C) x2 = 8(y + 3) (D) y2 = 8(x + 5) Sol. A line passing through the vertex (–3, 0) and perpendicular to directrix x + 5 = 0 is x-axis which is the axis of the parabola by definition. Let focus of the parabola is (a, 0). Since vertex, is the middle point of Z(–5, 0) and focus S, therefore 1 – a 2 ) 5 a ( 3       Focus = (–1,0) Thus the equation to the parabola is (x + 1)2 + y2 =(x + 5)2  y2 = 8(x +3) Ans. [A] Ex.3 The equation of the directrix of the parabola y2 = 12x is- (A) x + 3 = 0 (B) y + 3 = 0 (C) x – 3 = 0 (D) y – 3 = 0 Sol. Here a = 3, so the equation of the directrix is given by x = – a  x= – 3  x + 3 = 0 Ans. [A] Ex.4 The equation of the latus rectum of the parabola x2 = – 12y is- (A) y = 3 (B) x = 3 (C) y = – 3 (D) x = –3 Sol. Here a = 3 so the equation of the L.R. is given by y = – a  y = – 3 Ans. [C] 4. REDUCTION TO STANDARD EQUATION If the equation of a parabola is not in standard form and if it contains second degree term either in y or in x (but not in both) then it can be reduced into standard form. For this we change the given equation into the following forms- (y – k)2 = 4a (x – h) or (x – p)2 = 4b (y – q) And then we compare from the following table for the results related to parabola. Examples based on Reduction to Standard eqn. of a Parabola Ex.5 The axis of the parabola 9y2 –16x –12y –57 = 0 is (A) 3y = 2 (B) 2x = 3 (C) x +3y = 3 (D) y = 3 Equation of Vertex Axis Focus Directrix Equation of L.R. Length of L.R. Parabola (y– k)2 = 4a (x–h) (h,k) y = k (h + a, k) x + a – h = 0 x = a + h 4a (x–p)2 = 4b ( y–q) (p, q) x = p (p,b+ q) y + b – q = 0 y = b + q 4b
• 5. Sol. 9y2 –12y = 16x + 57  (3y – 2)2 = 16x + 57                 16 61 x 9 16 3 2 y 2 Which shows that equation of the axis is y –2/3 = 0 or 3y = 2 Ans. [A] Ex.6 Vertex, focus, latus rectum, length of the latus rectum and equation of directrix of the parabola y2 = 4x +4y are (A) (1, 2), (0, 2), y = 0, 4,x = – 2 (B) (–1, 2), (0, 2), x = 0, 4, x = –2 (C) (–1, 2), (1, 2), x = 0, 4, x = 2 (D) (–1, 2) (0, 2), y = 0, 2 , y = – 2 Sol. Given parabola y2 = 4x + 4y or (y – 2)2 =4(x +1) or Y2 = 4X Here X = x +1, Y= y – 2 vertex = (X = 0, Y = 0) or (x + 1 = 0, y – 2 = 0)  (–1, 2) Focus (X = 1, Y = 0) or (x + 1 = 1, y – 2 = 0)  (0, 2) Latus rectum  X = 1 x = 0 Length of Latus rectum = 4 equation of the directrix is X = –1  x +1 = –1  x = –2 Ans. [B] Ex.7 The vertex of the parabola x2 – 8y – x + 19 = 0 is- (A)       2 1 , 32 75 (B)       32 75 , 2 1 (C)        32 75 , 2 1 (D) None of these Sol. The given equation of Parabola can be written as 2 2 1 x        – 8y + 19 – 4 1 = 0 2 2 1 x        = 8y – 4 1 76   2 2 1 x        = 8        32 75 y  vertex =       32 75 , 2 1 Ans.[B] 5. GENERAL EQUATION OF A PARABOLA If (h,k) be the locus of a parabola and the equation of directrix is ax + by + c = 0, then its equation is given by 2 2 ) k y ( ) h x (    = 2 2 b a c by ax    which gives (bx– ay)2 + 2gx + 2fy + d = 0 where g, f, d are the constants. Note: (i) It is a second degree equation in x and y and the terms of second degree forms a perfect square and it contains at least one linear term. (ii) The general equation of second degree ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represents a parabola, if (a) h2 = ab (b)  = abc + 2fgh – af2 – bg2 – ch2  0 6. EQUATION OF PARABOLA WHEN ITS VERTEX AND FOCUS ARE GIVEN 6.1 If both lie on either of the coordinate axis : In this case first find distance 'a' between these points and taking vertex as the origin suppose the equation as y2 = 4ax or x2 = 4ay. Then shift the origin to the vertex. 6.2 When both do not lie on any coordinate axes In this case first find the coordinates of Z and equation of the directrix, then write the equation of the parabola by the definition. Examples based on Equation of Parabola when its vertex and focus are given Ex.8 If (0,4) and (0,2) are vertex and focus of parabola respectively, then its equation is- (A) y2 + 8x = 32 (B) y2 – 8x = 32 (C) x2 + 8y = 32 (D) x2 – 8y = 32 Sol. Since vertex and focus are on y-axis, so y-axis is the axis of the parabola. Distance between vertex and focus = 'a' = 2. So on taking vertex as origin, the equation of the parabola is x2 = – 4ay (negative because vertex lies above the focus) or x2 = – 8y. Now shifting the origin to its original position, the required equation will be x2 = –8(y – 4)  x2 + 8y = 32 Ans. [C]
• 6. 7. PARAMETRIC EQUATION OF PARABOLA The parametric equation of Parabola y2 = 4ax are x = at2, y = 2at Hence any point on this parabola is (at2, 2 at) which is called as 't' point. Note: (i) Parametric equation of the Parabola x2 = 4ay is x = 2at, y = at2 (ii) Any point on Parabola y2 = 4ax may also be written as (a/t2, 2 a/t) (iii) The ends of a double ordinate of a parabola can be taken as (at2, 2 at) and (at2, – 2at) (iv) Parametric equations of the parabola (y – h)2 = 4a (x – k)2 is x – k = at2 and y – h= 2 at Examples based on Parametric equation of Parabola Ex.9 The parametric equation of the curve y2 = 8x are- (A) x = 2t, y = 4t2 (B) x = 2t2, y = 4t (C) x = t2, y = 2t (D) None of these Sol. Here a = 2; y = 2at  y = 2.2.t = 4t x = at2  x = 2t2 Ans. [B] Ex.10 The parametric equation of the curve (y– 2)2 = 12 (x – 4) are- (A) 6t, 3t2 (B) 2 + 3t, 4 + t2 (C) 4 + 3t2,2 + 6t (D) None of these Sol. Here a = 3 y – 2 = 2at  y = 2 + 2.3t = 2 + 6t x – 4 = at2  x = 4 + 3.t2 = 4 + 3t2 Ans. [C] Ex.11 Parameter t of a point (4,–6) of the parabola y2 = 9x is- (A) 4/3 (B) – 4/3 (C) – 3/4 (D) – 4/5 Sol. Parametric coordinates of any point on parabola y2 = 4 ax are (at2, 2 at) Here 4 a = 9  a = 9/4  y coordinate 2 at = – 6  2 (9/4) t= – 6 t = – 4/3 Ans. [B] 8. CHORD 8.1 Equation of chord joining any two points of a parabola Let the points are (at1 2, 2 at1) and (at2 2, 2 at2) then equation of chord is- (y– 2at1) = 2 1 2 2 1 2 at at at 2 at 2   (x – at1 2)  y – 2at1 = 2 1 t t 2  (x – at1 2)  (t1 + t2) y = 2x + 2at1 t2 Note : (i) If 't1' and 't2' are the Parameters of the ends of a focal chord of the Parabola y2 = 4ax, then t1t2 = – 1 (ii) If one end of focal chord of parabola is (at2, 2at) , then other end will be (a/t2, – 2a/t) and length of focal chord = a (t + 1/t)2. (iii) The length of the chord joining two points 't1 ' and 't2' on the parabola y2 =4ax is a (t1 – t2) 4 ) t t ( 2 2 1   8.2 Length of intercept = ) mc a )( m 1 ( a m 4 2 2   Examples based on Chord Ex.12 If one extremity of the focal chord of the parabola y2 = 16x is (1,–4) then find other extremity. Sol. Here 2at = – 4 and a = 4  t = 4 . 2 4  = – 2 1 other extremity is         2 / 1 4 . 2 , 4 / 1 4 = (16,16) Ans. Ex.13 Find the length of a chord of a parabola y2 = 4x which passes through vertex and makes an angle of 45º with x- axis. Sol. Let the equation of chord is y = mx + c Here m = tan 45º = 1 as it passes through vertex (0,0) c = 0 Hence the equation is y = x and a = 1  = 4 ) 0 1 )( 1 1 ( 1   = 4 2 Ans. Ex.14 Length of intercept by the line 4y = 3x – 48 on the parabola y2 = 64x is-
• 7. (A) 1600 9 (B) 9 1600 (C) 9 160 (D) None of these Sol. 4y = 3x – 48  m = 3/4, c = –12 y2 = 64x  a = 16 Length of intercept = ) mc a ( ) m 1 ( a m 4 2 2   =                 4 3 12 16 16 9 1 16 16 9 4 = 9 1600 Ans. [B] 9. POSITION OF A POINT AND A LINE WITH RESPECT TO A PARABOLA 9.1 Position of a point with respect to a parabola A point (x1, y1) lies inside, on or outside of the region of the parabola y2 = 4ax according as y1 2 – 4ax1 < = or > 0 9.2 Line and Parabola : The line y = mx + c will intersect a parabola y2 = 4ax in two real and different, coincident or imaginary point, according as a – mc >, = < 0 Ex.15 For the parabola y2 = 8x point (2, 5) is (A) Inside the parabola (B) Focus (C) Outside the parabola (D) On the parabola Sol. (y2 – 8x)x = 2, y = 5 = (5)2 – 8 × 2 = 9 > 0  Point (2, 5) is outside parabola y2 = 8x Ans. [C] 10. TANGENT TO THE PARABOLA 10.1 Condition of Tangency : If the line y = mx + c touches a parabola y2 = 4ax then c = a/m Note: (i) The line y = mx + c touches parabola x2 = 4ay if c = – am2 (ii) The line x cos  + y sin  = p touches the parabola y2 = 4ax if a sin2  + p cos  = 0 (iii) If the equation of parabola is not in standard form, then for condition of tangency, first eliminate one variable quantity (x or y) between equations of straight line and parabola and then apply the condition B2 = 4AC for the quadratic equation so obtained. Examples based on Condition of tangency Ex.16 The line y = mx + 1 is a tangent to the parabola y2 = 4x if (A) m = 1 (B) m = 4 (C) m = 2 (D) m = 3 Sol. Here a = 1, c = 1, m = m. Now applying condition of tangency c = a/m, we have 1 = 1/m  m = 1 Ans. [A] Ex.17 If the line 2x – 3y = k touches the parabola y2 = 6x, then the value of k is (A) 27/4 (B) –81/4 (C) – 7 (D) – 27/4 Sol. Given 2 k y 3 x   ...(1) and y2 = 6x ...(2)          2 k y 3 6 y2  y2 = 3(3y + k)  y2 – 9y – 3k = 0 ...(3) If line (1) touches parabola (2) then roots of quadratic equation (3) is equal  (– 9)2 = 4 × 1 × (– 3k)  4 27 k   Ans. [D] 10.2Equation of Tangent 10.2.1 Point Form : The equation of tangent to the parabola y2 = 4ax at the point (x1, y1) is yy1 = 2a(x + x1) or T = 0 10.2.2 Parametric Form : The equation of the tangent to the parabola at t i.e. (at2, 2at) is ty = x + at2 10.2.3 Slope Form : The equation of the tangent of the parabola y2 = 4ax is y = mx + m a (at ,2at) 2 Note : (i) y = mx + a/m is a tangent to the parabola y2 = 4ax for all value of m and its point of contact is (a/m2, 2a/m). (ii) y = mx – am2 is a tangent to the parabola x2 = 4ay for all value of m and its point of contact is (2am, am2)
• 8. (iii) Point of intersection of tangents at points t1 and t2 of parabola is [at1t2, a(t1+t2)] (iv) Two perpendicular tangents of a parabola meet on its directrix. So the director circle of a parabola is its directrix or tangents drawn from any point on the directrix are always perpendicular. (v) The tangents drawn at the end points of a focal chord of a parabola are perpendicular and they meet at the directrix. Examples based on Equation of tangent Ex.18 The equation of the tangent to the parabola y2 = 4ax at the points (3, 2) is (A) 3y + x +3 = 0 (B) 3x + y +3 = 0 (C) 3x = y +3 (D) 3y = x +3 Sol. Since (3, 2) lies on the parabola y2 = 4ax, so 4 = 12a  a = 1/3 Now using T = 0, the equation of the tangent at (3, 2) is y(2) = 2a(x + 3)  y = 3 1 (x + 3)  3y = x + 3 Ans. [D] Ex.19 The equation of the tangent to the parabola y2 = 12x drawn at the upper end of its latus rectum is (A) y = x + 3 (B) x + y = 3 (C) y = x – 3 (D) None of these Sol. Here a = 3, so the upper end of LR = (3, 6). Hence the equation of the tangent at this point is y(6) = 6(x + 3)  y = x + 3 Ans. [A] Ex.20 The slope of tangent lines drawn from (3, 8) to the parabola y2 = –12x are (A) –3, –1/3 (B) 3, 1/3 (C) 3, –1/3 (D) –3, 1/3 Sol. Since 82 + 12 × 3  0, therefore the point (3, 8) is not on the parabola Now equation of any tangent to the parabola y2 = –12x is written as y = mx –       m 3 Since this line passes through (3, 8)  8 = 3m –       m 3  3m2 – 8m – 3 = 0 Solving this equation, we get m = 3, – 3 1 Ans. [C] Ex.21 If a tangent to the parabola y2 = ax makes an angle of 45º with x- axis, then its point of contact is- (A)       2 a , 4 a (B)       2 a , 2 a (C)        4 a , 2 a (D)        2 a , 4 a Sol. The slope of the tangent = tan 45º = 1  m = 1 and a  a/4  Point of contact =       1 4 / a . 2 , 1 4 / a 2 =       2 a , 4 a Ans. [A] Ex.22 Find the common tangent of the parabola x2 =4ay and y2 = 4ax(m > 0). Sol. Equation of tangent for x2 = 4ay is y = mx – am2 as this line also touches y2 = 4ax  – am2 = a/m (c = a/m)  m3 = – 1  m = – 1  equation of tangent  x + y + a = 0 Ans. 11. NORMAL TO THE PARABOLA 11.1 Equation of Normal : 11.1.1 Point Form : The equation to the normal at the point (x1, y1) of the parabola y2 = 4ax is given by y – y1 = a 2 y1  (x – x1) 11.1.2 Parametric Form : The equation to the normal at the point (at2,2at) is y + tx = 2at + at3 11.1.3 Slope Form : Equation of normal in terms of slope m is y = mx – 2am – am3. Note : (i) the foot of the normal is (am2, –2am) (ii) Condition for normal : The line y = mx + c is a normal to the parabola y2 = 4ax if c = –2am – am3 and x2 = 4ay if c = 2a + 2 m a (iii) Normal Chord : If the normal at 't1' meets the parabola y2 = 4ax again at the point ' t2' then this is called as normal chord. Again for normal chord t2 = – t1 – 1 t 2
• 9. (iv) If two normal drawn at point ' t1' and ' t2' meet on the parabola then t1t2 =2 Examples based on Equation and properties of Normal Ex.23 The equation of normal at the point (a/m2, 2a/m) on the parabola y2 = 4ax is- (A) m3y = m2x – 2am2 – a (B) y = mx – 2am – am3 (C) m3y = 2am2 – m2x + a (D) None of these Sol. The equation of the normal at (at2, 2at) of the parabola y2 = 4ax is written as y + tx = 2at + at3 For the given point t = 1/m, therefore equation of required normal is written as 3 m 1 a m 1 a 2 x m 1 y                       m3y = 2am2 – m2x + a Ans. [C] Ex.24 Find the equation of a normal at the parabola y2 = 4x which is parallel to the line y = 3x + 4. Sol. Let the equation is y = 3x + c This is normal to the parabola y2 = 4x then c = – 2 am – am3 = – 2.3 – 33 ( a = 1) = – 33 Hence equation is y = 3x – 33 Ans. Ex.25 If the line x + y = 1 is a normal to the parabola y2 = kx, then the value of k is Sol. As line y = –x + 1 is the normal to the parabola y2 = kx then 1 = – 2(a)(– 1) – a(– 1)3 a = 1/3  4a = k k = 4/3 Ans. Ex.26 Find the equation of a normal at the parabola y2 = 4x which passes through (3,0). Sol. Equation of Normal y = mx – 2 am – am3 Here a = 1 and it passes through (3,0) 0 = 3m – 2m – m3  m3 – m = 0  m = 0, ± 1 for m = 0  y = 0 m = 1  y = x – 3 m = –1  y = – x + 3 Ans. Ex.27 If the normal of the parabola y2 = 4ax drawn at (a, 2a) meets the parabola again at the point (at2, 2at) then t is equal to- (A) 3 (B) 1 (C) –1 (D) –3 Sol. If t' be the parameter of the given point, then 2at' = 2a  t' = 1 Now t= – t' – t 2   t = –1 – 2 1 = – 3 Ans. [D] Ex.28 Find the point where the normal at (4,4) meets the parabola y2 = 4x Sol. Here a = 1 and at1 2 = 4  t1 = 2 t2 = – 2 – 2 2 = – 3 Point is (at2 2, 2 at2) = {(–3)2,2.1.(–3)}= (9,– 6) Ans. Ex.29 If two of the normal of the parabola y2 = 4x, that pass through (15, 12) are 4x + y = 72, and 3x – y = 33, then the third normal is (A) y = x – 3 (B) x + y = 3 (C) y = x + 3 (D) None of these Sol. Here, If m1 , m2 , m3 are slopes of normal, then m1 +m2 +m3 =0 and m1 m2 m3 = a y1 a = 1 here m1 = –4, m2 = 3  – 4 + 3 + m3 = 0  m3 = 1 Also (–4) (3) (1) = – 1 12 is satisfied. But (15, 12) satisfies y = x – 3 Hence (A) is correct answer Ans. [A] Note: As seen value of m3 is not sufficient to locate the correct answer. The answer should be confirmed before it is reported.
• 10. Ex.30 If two normal drawn from any point to the parabola y2 = 4ax makes angle  and  with the axis such that tan .tan  = 2, then locus of this point is- (A) y2 = 4ax (B) x2 = 4ay (C) y2 = – 4ax (D) x2 = –4ay Sol. Let the point is (h, k). The equation of any normal to the parabola y2 = 4ax is y = mx – 2am – am3 passes through (h, k) k = mk – 2am – am3 am3 + m(2a – h) + k = 0 ...(i) m1 , m2 , m3 are roots of the equation then m1 .m2 .m3 = – a k but m1 m2 = 2, m3 = – a 2 k m3 is root of (i) a 3 a 2 k        – a 2 k (2a – h) + k = 0  k2 = 4ah Thus locus is y2 = 4ax Ans. [A] 12. PAIR OF TANGENTS If the point (x1, y1 ) is outside the parabola, then two tangents can be drawn. The equation of pair of tangents drawn to the parabola y2 = 4ax is given by SS1 = T2 i.e. (y2 – 4ax) (y1 2 – 4ax1) = [yy1 – 2a(x + x1)]2 12.1Locus of point of Intersection of tangents The locus of point of intersection of tangent to the parabola y2 = 4ax which are having an angle  between them is given by y2 – 4ax = (a+x)2 tan2  Note : (i) If  = 0º or  then locus is (y2 –4 ax) = 0 which is the given parabola. (ii) If  = 90º, then locus is x + a = 0 which is the directrix of the parabola. Ex.31 The equation of pair of tangents drawn from (1,4) to the parabola y2 =12x is- (A) 3x2 + y2 – 10 x + 4y – 3 = 0 (B) 3x2 + y2 – 10 x + 4xy + 4y – 3 = 0 (C) 3x2 + y2 + 10 x – 4xy – 4y + 3 = 0 (D) x2 + 3y2 + 10 x + 4xy – 4y – 3 = 0 Sol. From formula SS1 = T2 (y2 – 12x). 4 = (4y – 6x – 6)2  3x2 +y2 +10x– 4xy –4y+3 = 0 Ans. [C] Ex.32 Find the locus of perpendicular tangents of the parabola x2 – 4x + 2y + 3 = 0 Sol. The given equation of parabola can be written as x2 – 4x + 4– 4 + 2y + 3 = 0  (x–2)2 +2y – 1 = 0 (x–2)2 = – 2 (y –1/2) 4a = 2  a =1/2 Directrix  y – 1/2 = 1/2  y = 1 is the equation of locus. Ans. 13. CHORD OF CONTACT The equation of chord of contact of tangents drawn from a point (x1, y1) to the parabola y2 = 4ax is yy1 = 2a (x + x1) This equation is same as equation of the tangents at the point (x1, y1). Note : (i) The chord of contact joining the point of contact of two perpendicular tangents always passes through focus. (ii) Lengths of the chord of contact is a 1 ) a 4 y )( ax 4 y ( 2 2 1 1 2 1   (iii) Area of triangle formed by tangents drawn from (x1, y1) and their chord of contact is a 2 1 2 / 3 1 2 1 ) ax 4 y ( 
• 11. Examples based on Chord of contact Ex.33 The chord of contact drawn from (2,4) to the parabola y2 = 4x is- (A) 2y = x – 2 (B) y = 2x + 2 (C) y = 2x – 1 (D) 2y = x + 2 Sol. Here a = 1 equation of chord of contact is yy1= 2a (x + x1) 4y = 2 (x + 2)  2y = x + 2 Ans.[D] Ex.34 The area of triangle made by the chord of contact and tangents drawn from point (4,6) to the parabola y2 = 8x is- (A) 2 1 (B) 2 (C) 2 1 3 (D) None of these Sol. Area = a 2 1 (y1 2 – 4ax1)3/2 Here a = 2, (x1, y1) = (4,6)  Area of triangle = 2 . 2 1 (36– 32)3/2 = ( ) / 4 4 3 2 = 2 Ans. [B] Ex.35 The length of chord of contact of tangents drawn from point (–2, –1) to the parabola y2 = 4x is- (A) 2 2 (B) 3 5 (C) 8 (D) None of these Sol. Here a = 1, (x1, y1) (–2, –1) Hence length of chord of contact = 1 a (y 4ax )(y 4a ) 1 2 1 1 2 2   = (1 4)(1 4.1.( 2))    = 5 9 . = 3 5 Ans. [B] 14. EQUATION OFTHE CHORD WITH GIVEN MIDPOINT The equation of the chord at the parabola y2 = 4ax bisected at the point (x1, y1) is given by T = S1 i.e. yy1 – 2a(x + x1) = y1 2 – 4ax1 Ex.36 Find the equation of a chord of the parabola y2 = 6x if mid point of the chord is (– 1,1). Sol. Here a = 3/2 y = – 2 3 2 (x –1) = 1 – 6 (–1)  y – 3x + 3 = 7  y – 3 x = 4 Ans. 15. DIAMETER OF THE PARABOLA The locus of the mid points of a system of parallel chords of a parabola is called a diameter of the parabola. The equation of a system of parallel chord y = mx + c with respect to the parabola y2 = 4ax is given by y a m  2 Ex.37 The equation of system of parallel chords of the parabola y2 = 3 2 x is y + 2x + 1 = 0 then find its diameter. Sol. Here 4a = 3 2  a = 6 1 and m = – 2 Diameter is y a m  2  y = 2 6 1 . 2   y = – 6 1 This is the required equation of diameter. Ans. 15.1 Properties of Diameter : (i) Every diameter of a parabola is parallel to its axis. (ii) The tangents at the end point of a diameter is parallel to corresponding system of parallel chords. (iii) The tangents at the ends of any chord meet on the diameter which bisects the chord. 16.GEOMETRICAL PROPERTIES OF THE PARABOLA (i) The sub tangent at any point on the parabola is twice the abscissa or proportional to square of the ordinate of the point. (ii) The sub normal is constant for all points on the parabola and is equal to its semi latus rectum 2a.
• 12. (iii) The semi latus rectum of a parabola is the H.M. between the segments of any focal chord of a parabola i.e. if PQR is a focal chord, then QR PQ QR . PQ 2 a 2   (iv) The tangents at the extremities of any focal chord of a parabola intersect at right angles and their point of intersection lies on directrix i.e. the locus of the point of intersection of perpendicular tangents is directrix. (v) If the tangent and normal at any point P of parabola meet the axes in T and G respectively, then (a) ST = SG = SP (b) PSK is a right angle, where K is the point where the tangent at P meets the directrix. (c) The tangent at P is equally inclined to the axis and the focal distance. (vi) The locus of the point of intersection of the tangent at P and perpendicular from the focus on this tangent is the tangent at the vertex of the parabola. (vii) If a circle intersect a parabola in four points, then the sum of their ordinates is zero. (viii) The area of triangle formed inside the parabola y2 = 4ax is a 8 1 (y1 – y2)(y2 – y3) (y3 – y1) where y1,y2, y3 are ordinate of vertices of the triangle. (ix) The abscissa of point of intersection R of tangents at P(x1, y1) and Q(x2, y2) on the parabola is G.M. of abscissa of P and Q and ordinate of R is A.M. of ordinate of P and Q thus R        2 x x , x x 2 1 2 1 (x) If vertex and focus of a parabola are on the x-axis and at distance a and a' from origin respectively then equation of parabola y2 = 4(a' – a) (x – a) (xi) The area of triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points. Examples based on Geometrical properties of a parabola Ex.38 The latus rectum of a parabola whose focal chord is PSQ such that SP=3 and SQ= 2 is given by- (A) 24/5 (B) 6/5 (C) 12/5 (D) None of these Sol. Since the semi-latus rectum of a parabola is the HM of segments of a focal chord.  Semi-latus rectum = SQ SP SQ . SP 2  = 5 12  Latus rectum = 5 24 Ans. [A]
• 13. Review Chart for Standard Parabolas y2 = 4ax x2 = 4ay Diagram Vertex (A) (0, 0) (0, 0) Focus (S) (a, 0) (0,a) Axis y = 0 x = 0 Directrix x + a = 0 y + a = 0 Equation of LR x – a = 0 y – a = 0 Length of LR 4a 4a Extremities of LR(L1, L2) (a, 2a); (a, –2a) (2a, a) ; (–2a, a) Focal distance of (x, y) x + a y + a Parametric equations x = at2, y = 2at x = 2at, y = at2 Parametric points (at2, 2at) (2at, at2) Condition of tangency c = a/m c = –am2 (for y = mx + c) Tangent at (x1, y1) yy1 = 2a (x + x1) xx1 = 2a (y + y1) Tangent in slope form y = mx + a/m y = mx – am2 point of contact of above (a/m2, 2a/m) (2am, am2) Tangent at ' t' point ty = x + at2 tx = y + at2 Slope of tangent at ' t ' 1/t t Normal at (x1, y1) y – y1 =– 2a y1 (x – x1) y – y1 = 1 x a 2  (x – x1) Normal in slope form y = mx – 2am – am3 y = mx + 2a + a/m2 Foot of above normal (am2, –2am) (–2a/m, a/m2) Normal at ' t ' point y + tx = 2at + at3 ty + x = 2at + at3 Slope of normal at 't' –t – 1/t Condition of normal c = – 2am – am3 c = 2a + a/m2 (for y = mx + c) Director circle x + a = 0 y + a = 0 Diameter w.r.t. (y = mx + ) y = 2a/m x = 2am
• 14. Ex.1 The vertex of the parabola y2 +6x – 2y +13 =0 is (A) (1 , –1) (B) (–2, 1) (C) (3/2, 1) (D) (– 7/2,1) Sol. We have : y2 + 6x – 2y + 13 = 0  y2 – 2y = –6x – 13  (y – 1)2 = – 6(x + 2) Clearly, the vertex of this parabola is (–2, 1) Ans. [B] Ex.2 If vertex of parabola is (2, 0) and directrix is y- axis, then its focus is- (A) (2, 0) (B) (–2, 0) (C) (–4, 0) (D) (4, 0) Sol. Since the axis of the parabola is the line which passes through vertex and perpendicular to the directrix, therefore x-axis is the axis of the parabola. Obviously Z  (0, 0). Let focus of the parabola is S (a, 0). Since vertex (2,0) is mid point of ZS, therefore 2 0 a  = 2  a = 4.  Focus is (4, 0) Ans. [D] Ex.3 If the focus of a parabola is (1, 0) and its directrix is x + y = 5, then its vertex is- (A) (0, 1) (B) (0, –1) (C) (2, 1) (D) (3, 2) Sol. Since axis is a line perpendicular to directrix, so it will be x – y = k. It also passes from focus, therefore k = 1. So equation of axis is x – y = 1. Solving it with x + y = 5, we get Z  (3, 2). If vertex is (a, b), then a = 2, b = 1. Hence vertex is (2, 1). Ans. [C] Ex.4 The directrix and axis of the parabola 4y2 – 6x – 4y = 5 are respectively. (A) 8x + 11 = 0; y – 1 = 0 (B) 8x – 11 = 0, 2y – 1 = 0 (C) 8x + 11 = 0; 2y – 1 = 0 (D) None of these Sol. Here 4y2 – 4y = 6x + 5  4 2 2 1 y        = 6 (x + 1) Put y – 2 1 = Y, x + 1 = X The equation in standard form Y2 = 2 3 X 4a = 2 3  a = 8 3 Directrix, X + a = 0  x +1+ 8 3 = 0 8x + 11 = 0 Axis is Y = 0  y – 2 1 = 0 2y – 1 = 0 Ans. [C] Ex.5 The angle subtended by double ordinate of length 8a at the vertex of the parabola y2 = 4ax is- (A) 45º (B) 90º (C) 60º (D) 30º Sol. Let (x1 , y1 ) be any point on the parabola y2 = 4ax, then length of double ordinate 2y1 = 8a  y1 = 4a y1 2 = 4ax1  x1 = 4a  vertices of double ordinate are P(4a, 4a); Q(4a, –4a) If A is the vertex(0, 0), then Slope of AP = 1 = m1 Slope of AQ = –1 = m2  m1 m2 = –1  PAQ = 90º Ans. [B] Ex.6 The length of latus rectum of a parabola, whose focus is (2, 3) and directrix is the line x – 4y + 3 = 0 is- (A) 17 7 (B) 21 14 (C) 21 7 (D) 17 14 SOLVED EXAMPLES (3/8, 0) X O Y (–3/8, 0)
• 15. Sol. The length of latus rectum = 2 × perp. from focus to the directrix = 2 × 2 2 ) 4 ( ) 1 ( 3 ) 3 ( 4 2    = 17 14  The numerical length = 17 14 Ans. [D] Note:- The negative sign of the latus rectum may only be ignored if its length is asked. For other calculations it should be used. Ex.7 The coordinates of an endpoint of the latus rectum of the parabola (y – 1)2 = 4(x + 1) are (A) (0,–3)(B) (0,–1) (C) (0,1) (D) (1, 3) Sol. Shifting the origin at (–1, 1) we have        1 Y y 1 X x ...(i) Using (i), the given parabola becomes. Y2 = 4X The coordinates of the endpoints of latus rectum are (X = 1, Y = 2) and (X =1, Y= –2) Using (i), the coordinates of the end point of the latus rectum are (0,3) and (0, –1) Ans. [B] Ex.8 The locus of the middle points of all focal chords of the parabola y2 = 4ax is- (A) y2 = 2a(x + a) (B) y2 = 2ax ( C ) y2 = 2a (x – a) (D) None of these Sol. Let the middle point be (h, k), then equation of the chord is given by k2 – 4ah = yk – 2a(x + h) Since it passes through (a, 0),  k2 – 4ah = – 2a(h + a)  k2 = 2a(h – a) Hence required locus is y2 = 2a(x – a) which is parabola. Ans. [C] Ex.9 The length of the chord of parabola x2 = 4ay passing through the vertex and having slope tan  is- (A) 4a cosec  cot  (B) 4a tan  sec  ( C ) 4a cos  cot  (D) 4a sin  tan  Sol. Let A be the vertex and AP be a chord of x2 = 4ay such that slope of AP is tan . Let the coordinates of P be(2at, at2 ) Then, Slope of AP = at 2 at2 = 2 t tan = 2 t   t 2tan Now, AP = 2 2 2 ) 0 at ( ) 0 at 2 (    = at 2 t 4  = 2a tan  a tan 4 4 2  = 4a tan  sec  Ans. [B] Ex.10 The point on y2 = 4ax nearest to the focus has its abscissa equal to (A) –a (B) a (C) a/2 (D) 0 Sol. Let P(at2 , 2at) be a point on the parabola y2 = 4ax and S be the focus of the parabola. Then, SP = a + at2 [ focal distance = x + a] Clearly, SP is least for t = 0. Hence, the abscissa of P is at2 = a × 0 = 0 Ans. [D] Ex.11 Tangents are drawn to the points of intersection of the line 7y – 4x = 10 and parabola y2 = 4x,.then the point of intersection of these tangents is (A)       2 5 , 5 7 (B)       2 7 , 2 5 (C)         2 7 , 2 5 (D)         2 5 , 2 7 Sol. Here let the point be (x1 , y1 )  chord of contact yy1 = 2 (x + x1 ) compare with the given line 10 x 2 1  = 7 y1  = 4 2   x1 = 5/2, y1 = 7/2 Point reqd.       2 7 , 2 5 Ans. [B] Ex.12 If the line x + my + am2 = 0 touches the parabola y2 = 4ax, then the point of contact is (A) (am2 , – 2am) (B)        m a 2 , m a 2 (C) (–am2 , – 2am) (D) The line does not touch. Sol. Here x = –my – am2
• 16. Putting in y2 = 4ax, we get y2 = 4a (–my – am2 ) or y2 + 4amy + 4a2 m2 = 0 or (y + 2am)2 = 0 Which is a perfect square, hence the line is a tangent  y = –2am, also x = am2 the point reqd. is (am2 , –2am) Ans. [A] Note:- For condition of tangency, solve the equation of line and curve and put B2 –4AC = 0 for the quadratic equation obtained by the two, (the line and curve equation) Ex.13 The common tangent of the parabola y2 = 8ax and the circle x2 + y2 = 2a2 is (A) y = x + a (B) y = x – a (C) y = x –2a (D) y = x + 2a Sol. Any tangent to parabola is y = mx + m a 2 Solving with the circle x2 + (mx + m a 2 )2 = 2a2 B2 – 4AC = 0 gives m = ± 1 Otherwise Perp. from (0, 0) = radius a 2  2 m 1 m a 2        = a 2  m = ± 1 Tangent y = ± x ± 2a  y = x + 2a is correct option. Ans. [D] Ex.14 If two parabolas y2 = 4x and x2 = 32y intersect in the point (16, 8) at an angle , then  is equals- (A) tan–1 (3/5) (B)  (C) tan–1 (4/5) (D) /2 Sol. The slope of tangent at the point (16, 8) to the parabola y2 = 4x is 2 1 a y = 4 1 and slope of tangent to the parabola x2 = 32y is 1.  tan = ) 4 / 1 ( 1 ) 4 / 1 ( 1   = 3 5 Ans. [A] Ex.15 The slope of tangents drawn from a point (4, 10) to the parabola y2 = 9x are- (A) 4 3 , 4 1 (B) 4 9 , 4 1 (C) 3 1 , 4 1 (D) None of these Sol. The equation of a tangent of slope m to the parabola y2 = 9x is y = mx + m 4 9 If it passes through (4,10), then 10 = 4m + m 4 9  16m2 – 40m + 9 = 0 (4m –1)(4m –9) = 0  m = 4 1 , 4 9 Ans. [B] Ex.16 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) 2 (C) 1/3 (D) 1/2 Sol. Any tangent to y2 = 4x is y = mx + 1/m If it is drawn from (–2, –1), then –1 = –2m + 1/m  2m2 – m – 1 = 0 If m = m1 , m2 then m1 + m2 =1/2, m1 m2 =–1/2  tan a = 2 1 2 1 m m 1 m m   = 2 1 2 1 2 2 1 m m 1 m m 4 ) m m (    2 / 1 1 2 4 / 1   = 3 Ans. [A] Ex.17 If the straight line x + y = 1 is a normal to the parabola x2 = ay, then the value of a is (A) 4/3 (B) 1/2 (D) 3/4 (D) 1/4 Sol. We know that equation of normal to the parabola x2 = ay is y =mx + 4 a 2 + 2 m 4 a Given that x + y = 1 or y = –x + 1 is normal to the parabola therefore m = –1 and 2 a + 2 m 4 a = 1  2 a + 4 a = 1 4 a 3 = 1 a = 3 4 Ans. [A]
• 17. Ex.18 If line y = 2x + k is normal to the parabola y2 = 4x at the point (t2 , 2t), then- (A) k = –12, t = –2 (B) k = 12, t = –2 (C) k = 12, t = 2 (D) None of these Sol. Since normal to the parabola y2 = 4x at (t2 , 2t) is y + tx = 2t + t3 . Comparing it with y = 2x + k, we get t = –2, k = 2t + t3 = –12 Ans. [A] Ex.19 Which of the following lines, is a normal to the parabola y2 = 16x (A) y = x – 11 cos – 3 cos 3 (B) y = x – 11 cos – cos 3 (C) y = (x – 11) cos + cos 3 (D) y = (x – 11) cos – cos 3 Sol. Here a = 4 condition of normality c = –2am – am3 (1) and (2) are not clearly the answer as m = 1 for (3), (4) m = cos c = –2(4) cos – 4 cos3  = – 8 cos – (3 cos + cos 3) = –11 cos – cos 3 Hence (D) is correct Ans. [D] Ex.20 A normal is drawn to the parabola y2 = 4ax at the point (2a, –2 2 a) then the length of the normal chord, is (A) 4 2 a (B) 6 2 a (C) 4 3 a (D) 6 3 a Sol. Here comparing (2a,–2 2 a) with (am2 , –2am) we get m = 2 Now length of normal chord = 2 m a 4 (1 +m2 )3/2 = 2 a 4 (1 + 2)3/2 = 2a 3 3 = 6 3 a Ans. [D] Ex.21 If a normal chord of the parabola y2 =4ax subtend a right angle at the vertex, its slope is- (A) ± 1 (B) ± 2 (C) ± 3 (D) None of these Sol. If P(at1 2 , 2at1 ) be one end of the normal, the other say Q(at2 2 , 2at2 ) then t2 = – t1 – 1 t 2 ...(1) Again slope of OP = 2 1 1 at at 2 = 1 t 2 Slope of OQ = 2 t 2  1 t 2 × 2 t 2 = –1  t1 t2 = –4 ...(2) From (1) and (2) – 1 t 4 = –t1 – 1 t 2  + 1 t 2 = t1  t1 2 = 2  t1 = ± 2 Ans. [B] Note:- Slope of the normal at P is (–t1 ) Ex.22 If the tangents at P and Q on a parabola (whose focus is S) meet in the point T, then SP, ST and SQ are in- (A) H.P. (B) G.P. (C) A.P. (D) None of these Sol. Let P (at1 2 , 2at1 ) and Q (at2 2 , 2at2 ) be any two points on the parabola y2 = 4ax, then point of intersection of tangents at P and Q will be T  [at1 t2 ,a(t1 + t2 )] Now SP= a(t1 2 + 1) SQ = a(t2 2 + 1) ST =a ) 1 t )( 1 t ( 2 2 2 1    ST2 = SP.SQ  SP, ST and SQ are in G.P. Ans. [B] Ex.23 If the distance of 2 points P and Q from the focus of a parabola y2 = 4ax are 4and 9 respectively, then the distance of the point of intersection of tangents at P and Q from the focus is (A) 8 (B) 6 (C) 5 (D) 13 O 90° Q P X Y
• 18. Sol. If S is the focus of the parabola and T is the point of intersection of tangents at P and Q, then ST2 = SP × SQ  ST2 = 4 × 9  ST = 6 Ans. [B] Ex.24 From the point (–1, 2) tangents are drawn to parabola y2 =4x, then the area of the triangle formed by the chord of contact and the tangent is- (A) 2 2 (B) 3 2 (C) 4 2 (D) 8 2 Sol. Here chord of contact y(2) = 2(1) (x –1) or y = x – 1 Length of the chord = 2 1 4 ) 1 1 )( 1 1 ( 1   = 8 Area required = 2 1 × 8 × (perpendicular from (–1, 2) to chord) = 4 × 1 1 1 1 2    = 8 2 Ans. [D] Note:- Direct formula may be used.