# Parabola-02-Solved Example

Educator em Study Innovations
24 de May de 2023
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### Parabola-02-Solved Example

• 1. SOLVED SUBJECTIVE EXAMPLES Example 1 : Show that the line x cos + ysin = p touches the parabola y2 = 4ax if p cos + asin2  = 0 and that the point ofcontact is (a tan2 , – 2a tan) Solution : The givenline is x cos + y sin = p or y = – x cot  + p coses   m = – cot  and c = p cosec  since the givenline touches the parabola  c = m a or cm = a or (p cosec ) (– cot  = a and point ofcontact is       m a 2 , m a 2 i.e.          cot a 2 , cot a 2 or (atan2 , – 2atan) Example 2 : Prove that the line  x + m y = 1 touches the parabola y2 = 4a (x + b) if m2 ( + b) + a2 = 0. Solution : The givenparabola is y2 = 4a (x + b) .......(1) Vertex ofthis parabola is (– b, 0) Now shifting (0, 0) at (– b, 0) then then x = X + (– b) and y = Y + 0 ........(2) or x + b = X and y = Y ........(3) From (1), Y2 = 4aX and the line  x + m y = 1 reduces to  b X  + m Y = 1 or Y = m          b X 1
• 2.  Y =         m X + m         b 1 .......(4) The line (4) willtouch the parabola (3), if m         b 1 =         m a or  2 m         b 1 = – a or m2 ( + b) + a2 = 0 Example 3 : Find the equations of the straight lines touching both x2 + y2 = 2a2 and y2 = 8ax. Solution : The givencurves are x2 + y2 = 2a2 and y2 = 8ax The parabola (2) is y2 = 8ax or y2 = 4(2a)x  Equation oftangent of(2) is y = mx + m a 2 or m2 x – my + 2a = 0 ......(3) It is also tangent of(1), then the lengthofperpendicular fromcentre of(1) i.e. (0, 0)to (3) must be equalto the radius of(1) i..e., 2 a .   2 2 2 m ) m ( a 2 0 0     = a 2 or 2 2 4 2 a 2 m m a 4   or m4 + m2 – 2 = 0 or (m2 + 2) (m2 – 1) = 0  m2 + 2  0 (gives theimaginaryvalues)  m2 – 1 = 0  m = ± 1 Hence from(3) the required tangents are x ± y + 2a = 0.
• 3. Example 4 : Tangents are drawnfromthe point (x1 , y1 ) to the parabola y2 = 4ax, show that the length oftheir chord ofcontact is | a | 1 ) a 4 y )( ax 4 y ( 2 2 1 1 2 1   . Solution : Given parabola is y2 = 4ax ......(1) Let P  (x1 , y1 ) Let the tangent fromP touch the parabola at Q (at1 2 , 2at1 ) and R(at2 2 , 2at2 ) then P is the point of intersectionoftangents  x1 = at1 t2 & y1 = a(t1 + t2 ) or t1 t2 = a x1 and t1 + t2 = a y1 Now QR = 2 2 1 2 2 2 2 1 ) at 2 at 2 ( ) at at (    = ] 4 ) t t [( ) t t ( a 2 2 1 2 2 1 2    = |a| |t1 – t2 | } 4 ) t t {( 2 2 1   = | a |          a x 4 a y 1 2 2 1          4 a y 2 2 1 {fromequation2} = |a| | a | ) ax 4 y ( 1 2 1  . | a | ) a 4 y ( 2 2 1  P (x , y ) 1 1 R Q Chord of Contact (at , 2at ) 1 1 2 (at , 2at ) 2 2 2 = | a | 1 ) a 4 y )( ax 4 y ( 2 2 1 1 2 1   . Example 5 : A ray of light is coming along the line y = b from the positive direction of x-axis and strikes a concave mirror whose intersectionwith the x-yplane is a parabola y2 = 4ax. Find the equation of the reflected ray and show that it passes through the focus of the parabola. Both a and b are positive. Solution : A rayoflight along y= b intersects the parabola at         b , a 4 b P 2 Equation to the parabola is y2 = 4ax ... (1) Differentiating (1) dx dy = y a 2
• 4. Slope of normalat P is a 2 b   Ifthe angle between rayoflight and the normalis  then tan = a 2 b Let PQ be reflected ray Angle between PQ and the normalat P is also  Let the slope of PQ be m tan = m a 2 b 1 a 2 b m a 2 b m a 2 b 1 m a 2 b         a 2 b m m a 4 b a 2 b 2 2     X Normal Y S A y=b P         b , a 4 b2  a 2 b 2 a 4 b a 4 m 2 2 2             m = – 2 2 b a 4 ab 4   Equation to the reflected ray is y– b = –           a 4 b x b a 4 ab 4 2 2 2              a 4 b ax 4 b a 4 ab 4 b y 2 2 2 4abx + (5a2 – b2 ) y – 4a2 b = 0 Substitute S(a, 0)  Reflected raypasses through the focus. Example 6 : Find theequationofaxis, vertex, directrix, extremities oflatus rectumand lengthof latus rectumof the parabola x2 – 3y + 2x + 3 = 0. Solution : x2 –3y + 2x + 3 = 0  x2 + 2x + 1 = 3y – 2  (x + 1)2 = 3        3 2 y . Its axis is the line x + 1 = 0. It’s vertex is        3 2 , 1 . It’s focus is x + 1 = 0, y– 3 2 = 4 3 i.e.,        12 17 , 1 . It’s directrix is the line y– 3 2 = – 4 3 , i.e., y+ 12 1 = 0. It’s latus rectumis the line y– 3 2 = 4 3 . i.e., y– 12 17 = 0. Extremities ofit’s latus
• 5. rectum are x+ 1 = 2 3 , y– 3 2 , i.e.,       12 17 , 2 1 . Lengthof it’s latus rectumis 3 units. Example 7 : Ifthe line ax+ by+ c = 0 is a tangent to the parabola y2 = 4ax thenfind thecorresponding point of contact. If the line ax + by + c = 0 is a normal to the parabola y2 = 4ax then find the foot of its normal. Solution : Let the point of contact of tangent be P1 (at1 2 , 2at1 ). Lines ax + by + c = 0 and yt1 – x – at1 2 = 0 must be identical,  b t1 = a 1  = c at 2 1   t1 = – a b . Thus P1           b 2 , a b2 . If foot of normal be P2 (at2 2 , 2at2 ) then y + t2 x – 2at2 – at2 3 = 0 and ax + by + c = 0 must be identical,  b 1 = a t2 = c ) at at 2 ( 3 2 2    t2 = b a Thus P2          b a 2 , b a 2 2 3 . Example 8 : Prove that the area of the triangle formed by three points on a parabola is twice the area of the triangle formedbythe tangents at these points. Solution : Let the three points on the parabola be at ( 1 2 1 at 2 , at ), 2 2 at ( , 2at2 ), (at3 2 , 2at3 ). The area of the  formed bythese points 1 at 2 at 1 at 2 at 1 at 2 at | 2 1 3 2 3 2 2 2 1 2 1  | = |a2 (t1 – t2 ) (t2 – t3 ) (t3 – t1 )| The intersectionofthe tangents, at these points, are the points {at1 t2 , a(t1 + t2 )}, {at2 t3 , a(t2 + t3 )}, {at3 t1 , a(t3 + t1 )} The area of the  formed bythese points = 2 1 |a2 (t1 – t2 ) (t2 – t3 ) (t3 – t1 )|.
• 6. Example 9 : Find thelocus ofthe point ofintersectionoftwo normals to a parabolawhichare at right angles to one another. Solution : The equationofthe normal to the parabola y2 = 4ax is y = mx – 2am – am3 It passesthrough the point (h, k) if k = mh – 2am – am3  am3 + m(2a - h) + k = 0 ... (1) Let the roots ofthe above equation be m1 , m2 and m3 . Let the perpendicular normals correspond to the values ofm1 and m2 so that m1 m2 = –1. From the equation (1), m1 m2 m3 = a k  Since m1 m2 = –1, m3 = a k Since m3 is a root of (1), we have a a k a k 3        (2a - h) + k = 0  k2 + a(2a - h) + a2 = 0  k2 = a(h – 3a) Hence the locus of(h, k) is y2 = a(x– 3a). Example 10 : Three normals froma point to the parabola y2 = 4ax meet the axisofthe parabola inpoints whose abscissa are inA.P. Find the locus of the point. Solution : The equationofanynormalto the parabola is y = mx – 2am – am3 It passesthrough the point (h, k) if am3 + m (2a - h) + k = 0 ... (1) The normalcuts the axis ofthe parabola viz., y= 0 at point where x = 2a + am2 Hence the abscissaofthe points inwhich the normals through(h, k) meet the axisofthe parabola are x1 = 2a + am1 2 , x2 = 2a + am2 2 , x3 = 2a + am3 2 Since x1 , x2 , x3 are inA.P. (2a + 2 1 am ) + (2a + 2 3 am ) = 2 (2a + 2 2 am )  2 2 2 2 2 1 m 2 m m   ... (2) Also, from (1), m1 + m2 + m3 = 0, ... (3) m2 m3 + m3 m1 + m1 m2 = a h a 2  ... (4)
• 7. and m1 m2 m3 = – a k ... (5) From(3), (m1 + m3 )2 = 2 2 m  2 3 2 1 m m  + 2m1 m3 = 2 2 m  2 2 2 2 2 2 2 2 am k 2 m m am k . 2 m 2      k 2 am3 2  Since m2 is a root of (1), am2 3 + m2 (2a – h) + k = 0  2k + m2 (2a – h) + k = 0  {m2 (h–2a)}3 = 27k3  a k 2 (h – 2a)3 = 27k3  27 ak2 = 2 (h – 2a)3 . Hence the locus of (h, k) is 27 ay2 = 2(x – 2a)3 . Example 11 : P, Q are the points t1 , t2 on the parabola y2 = 4ax. The normals at P, Q meet on the parabola. Show that the middle point of PQ lies on the parabola y2 = 2a(x + 2a). Solution : The points P and Q are (at1 2 , 2at1 ) and (at2 2 , 2at2 ) As the normals at t1 & t2 meet onthe parabola, t1 t2 = 2 ... (i) Also if (x1 , y1 ) be the midpoint ofPQ, then x1 = 2 1 (at1 2 + at2 2 ) and y1 = 2 1 (2at1 + 2at2 ) ... (ii) From (iii) we get (t1 + t2 )2 = (y1 /a)2  (y1 /a)2 = t1 2 + t2 2 + 2t1 t2 = (2x1 /a) + 4, using (i) and (ii)  y1 2 = 2a (x1 + 2a) Hence the locus of (x1 , y1 ) is y2 = 2a(x + 2a). Example 12 : Ifa circle intersects the parabola y2 = 4axin four points thenshow that the algebraic sumofthe ordinatesiszero.Also show that theline joiningonepair ofthese fourpoints andthe linejoiningthe other pair have slopes equalinmagnitude. Solution : Solving y2 = 4ax and x2 + y2 + 2gx + 2fy+ c = 0 we have y1 + y2 + y3 + y4 = 0 (y1 , y2 , y3 , y4 are ordinates of the points ofintersection)  y1 + y2 = – (y3 + y4 ) Slope ofline joining (x1 , y1 ) and (x2 , y2 ) is m(say)  2 1 2 1 2 2 1 2 1 2 1 2 y y a 4 a 4 y a 4 y y y x x y y m         Slope ofline joining (x3 y3 ) and (x4 , y4 ) is  m (say)
• 8.  4 3 2 3 2 4 3 4 3 4 3 4 y y a 4 a 4 y a 4 y y y x x y y m          Since y1 + y2 = –(y3 + y4 ), We have m = –  m  Slopes are equalin magnitude. Example 13 : Tangents aredrawn to the parabolaat three distinct points. Prove that these tangent lines always make a triangle and that the locusofthe orthocentre ofthe triangle is the directrixofthe parabola. Solution : Let the three points beA(at1 2 , 2at1 ), B(at2 2 , 2at2 ), C(at3 2 , 2at3 ). Tangents at these point are t1 y = x + at1 2 , t2 y = x + at2 2 , t3 y = x + at3 2 . Since t1 , t2 , t3 are distinct, no two tangents are parallel or coincident. Hence these tangents will form a triangle. The vertices of the triangle are [at1 t2 , a(t1 + t2 )], [at2 t3 , a(t2 + t3 )] and [at3 t1 , a(t3 + t1 )]. Equation ofthe two altitudes are [y – a(t2 + t3 )] = –t1 [x - at2 t3 ] ... (1) and [y – a (t3 + t1 )] = –t2 [x – at3 t1 ] ... (2) Subtracting (2) from(1), we get x = –a Hence the locus ofthe orthocentere is x + a = 0 which is the directrixofthe parabola. Example 14 : Tangents are drawn to a parabola from a point T. IfP, Q are the points of contact, prove that the perpendicular distances fromP, T and Q upon anytangent to the parabola are in G.P. Solution : Let the equation ofthe parabola be y2 = 4ax. Let P(at1 2 , 2at1 ) and Q(at2 2 , 2at2 ) be the two points ofcontact. Then the coordinatesofT willbe (at1 t2 , a(t1 + t2 )). Theequation ofthe tangent to the parabola at anypoint (at2 , 2at) is ty= x+ at2 . Let p1 , p2 , p3 be the length ofthe perpendicular on this tangent fromthe points P, T and Q respectively. Then | t 1 ) t t ( a | | t 1 att 2 at at | p 2 2 1 2 1 2 2 1 1        | t 1 ) t t ( a | | t 1 att 2 at at | p 2 2 2 2 2 2 2 2 3        so that | ) t 1 ( ) t t ( ) t t ( a | p p 2 2 2 2 1 2 3 1     = 2 2 2 1 2 2 1 t 1 ) t t ( at at t at           = p2 2  p1 , p2 , p3 are in G.P. Example 15 : Three normals aredrawnfromthe point (c, 0)to the curve y2 =x. Show that c must begreater than 1/2. One normalis always the x-axis. Find c for whichthe other two normalsare perpendicular to eachother.
• 9. Solution : Equation to normalto the parabola y2 = x is y = mx – 2 m – 4 m3 is passing through (c, 0) 0 = cm – 4 m 2 1 c m 4 m 2 m 3 3           m = 0 or 4 m 2 1 c 2   0 2 1 c            0 4 m ce sin 2 c  2 1 . If c = 2 1 , then m = 0 Then onlyone normalwillbethere i.e., x-axis. Since three normals are there,  c > 2 1 Then                    2 1 c 2 2 1 c 2 = –1  c – 4 1 2 1   c = 4 3
• 10. SOLVED OBJECTIVE EXAMPLES Example 1 : Equation ofparabola having it’s focus at S(2, 0) and one extremityofit’slatus rectumas (2, 3) is (A) y2 = 4(3 – x) (B) y2 = 4(1 – x) (C) y2 = 8(3 – x) (D) y2 = 8(1 – x) Solution : Clearlythe other extremityoflatus rectumis (2, – 2). It’s axis is x-axis. Corresponding value of a = 2 0 2 =1. Hence it’s vertexis (1, 0) or(3, 0). Thus it’sequationis y2 =4(x– 1) or y2 = – 4(x – 3). Example 2 : Equation ofparabola having the extremities ofit’s latus rectumas (3, 4) and (4, 3) is (A) 2 3 7 x        + 2 3 7 y        = 2 2 6 y x         (B) 2 2 7 x        + 2 2 7 y        = 2 2 8 y x         (C) 2 2 7 x        + 2 2 7 y        = 2 2 4 y x         (D) None ofthese Solution : Focus is       2 7 , 2 7 and it’s axis is the line y= x. Corresponding value of‘a’is 4 1 ) 1 1 (  = 4 2 . Let the equation of it’s directrix be y + x + = 0.  2 | 4 x 3 |    = 2. 4 2  = – 6, – 8 Thus equationofparabola is 2 2 7 x        + 2 2 7 y        = 2 ) 6 y x ( 2   or 2 2 7 x        + 2 2 7 y        = 2 ) 8 y x ( 2   . Example 3 : An equilateral triangle is inscribed in the parabola y2 = 4ax, such that one vertex ofthis triangle coincides with the vertexofthe parabola. Side length ofthistriangle is - (A) 4a 3 (B) 6a 3 (C) 2a 3 (D) 8a 3
• 11. Solution : If triangle OAB is equilateral then OA = OB = AB. Thus AB will be a double ordinate of the parabola. Thus AOX = XOB = 6  . Let A = (at1 2 , 2at1 ) then B  (at1 2 , 2at1 ) mOA = 1 t 2 = 3 1  t1 = 3 2  AB = 4at1 = 8a 3 units. Example 4 : If the normal drawn to parabola y2 = 4ax at the point A(at1 2 , 2at1 ) meets the curve again at B(at2 2 , 2at2 ) then (A) |t2 |  2 2 (B) |t2 |  2 2 (C) |t1 |  2 2 (D) |t1 |  2 2 Solution : We have t2 = – t1 – 1 t 2  t1 2 + t1 t2 + 2 = 0. Since ‘t1 ’ is real, thus t2 2 – 8  0  |t2 |  2 2 . Example 5 : Locus oftrisectionpoint ofanydouble ordinate ofy2 = 4ax is (A) 3y2 = 4ax (B) y2 = 6ax (C) 9y2 = 4ax (D) None ofthese Solution : LetAB be a double ordinate, whereA  (at2 , 2at), B  (at2 , –at). IfP(h, k) be it’s trisection point then 3h = 2at2 + at2 , 3k =4at – 2at  t2 = a h , t = a 2 k 3 Thus locus is 2 2 a 4 k 9 = a h , i.e., 9y2 = 4ax. Example 6 : Tangents drawn to parabola y2 = 4ax at the pointAand B intersect at C. If‘S’be the focus ofthe parabola then SA, SC and SB are in (A)A.P. (B) G.P. (C) H.P. (D) noneofthese
• 12. Solution : IfA  (at1 2 , 2at1 ), B  (at2 2 , 2at2 ),  C  (at1 at2 , a(t1 + t2 )). Now SA = a + at1 2 , SB = a + at2 2 , SC = 2 2 1 2 2 2 1 ) t t ( a ) a t at (     a 2 2 2 1 2 2 2 1 t t 1 t t    = a ) t 1 )( t 1 ( 2 2 2 1   Clearly SC2 = SA. SB Example 7 : Double ordinatesAB of the parabola y2 = 4axsubtends an angle /2 at the focus ofthe parabola then tangents drawnto parabola atAand B willintersect at (A) (– 4a, 0) (B) ( –2a, 0) (C) (–3a, 0) (D) None ofthese Solution : Let A  (at2 , 2at), B  (at2 , – 2at). mOA = t 2 , mOB = – t 2 . Thus t 2 . – t 2 = – 1  t2 = 4. Thus, tangentswillintersect at (–4a, 0). Example 8 : Ifa normal chord ofy2 = 4ax subtends anangle /2 at the vertex ofthe parabola thenit’s slope is equalto (A) ± 1 (B) ± 2 (C) ± 2 (D) noneofthese Solution : Let AB be a normal chord, whereA (at1 2 , 2at1 ) and B  (at2 2 , 2at2 ). We have t2 = – t1 – 1 t 2 and t1 t2 = 4  t1 t2 = – t1 2 – 2 = – 4  t1 2 = 2. Now slope of chord AB = 2 1 t t 2  = – t1 = ± 2 Example 9 : Length ofthe shortest normalchord ofthe parabola y2 = 4ax is (A) a 27 (B) 3a 3 (C) 2a 27 (D) noneofthese Solution : Let AB be a normal chord where A (at1 2 , 2at1 ), B  (at2 2 , 2at2 ). We have t2 = – t1 – 1 t 2 . AB2 = [a2 (t1 – t2 )2 (t1 + t2 )2 + 4] = a2 2 1 1 1 t 2 t t                   4 t 4 2 1 = 4 1 3 2 1 2 t ) t 1 ( a 16 
• 13.  1 2 dt ) AB ( d = 16a2            8 1 3 1 3 2 1 1 2 2 1 4 1 t t 4 . ) t 1 ( ] t 2 . ) t 1 ( 3 [ t = 5 1 2 2 1 2 t ) t 1 ( 32 . a  (t1 2 – 2) t1 = 2 is indeed the point ofminima ofAB2 . Thus ABmini = 2 a 4 (1 + 2)3/2 = 2a 27 units. Example 10 : Ifthree distinct and real normals can be drawn to y2 = 8x fromthe point (a, 0) then (A) a > 2 (B) a > 4 (C) a (2, 4) (D) noneofthese Solution : Equation of normal in terms of m is y = mx – 4m – 2m3 . If it passes through (a, 0) then am – 4m – 2m3 = 0  m(a – 4 – 2m2 ) = 0  m = 0, m2 = 2 4 a  . For three distinct normal, a – 4 > 0  a > 4 Example 11 : An equilateraltriangle SAB is inscribed inthe parabola y2 = 4axhaving it’s focus at ‘S’. Ifchord AB lies towards the left ofS, thenside length ofthis triangle is (A) 3a(2 – 3 ) (B) 4a(2 – 3 ) (C) 2(2 – 3 ) (D) 8a(2 – 3 ) Solution : Let A (at1 2 , 2at1 ), B  (at2 2 , – 2at1 ). We have mAS = tan        6  a at at 2 2 1 1  = – 3 1  t1 2 + 2 3 t1 = – 1 = 0  t1 = – 3 ± 2 Clearly t1 = – 3 – 2 is rejected. Thus t1 = (2 – 3 ). HenceAB = 4at1 = 4a(2 – 3 ). Example 12 : Locus of the midpoint of anyfocalchord ofy2 = 4ax is (A) y2 = a(x – 2a) (B) y2 = 2a(x – 2a) (C) y2 = 2a(x – a) (D) noneofthese
• 14. Solution : Let the midpoint be P(h, k). Equationofthis chord isT = S1 . i.e., yk – 2a(x +h) = k2 –4ah. It must pass through(a, 0)  2a(a + h) = k2 – 4ah. Thus required locus is y2 = 2ax – 2a2 . Example 13 : Slope of the normal chord ofy2 = 8xthat gets bisected at (8, 2) is (A) 1 (B) – 1 (C) 2 (D) – 2 Solution : Let AB be the normal chord where A (2t1 2 , 4t1 ), B  (2t2 2 , 4t2 ). It’s slope = 2 1 t t 2  We also have t2 = – t1 – 1 t 2 and 16 = 2(t1 2 + t2 2 ), 4 = 4(t1 + t2 )  t1 + t2 = 1. Thus slope is 2. Example 14 : The lengthoflatus rectumofhe parabola, whosefocus is (a sin2, a cos2) anddirectrixisthe line y = a, is (A) |ra cos2 | (B) |4a sin2 | (C) |4a cos2| (D) |a cos2| Solution : Distance offocus fromdirectrix is |a cos 2 – a|. Thus lengthoflatus rectumis|4a sin2 |. Example 15 : In the adjacent figure a parabola is drawn to pass through the vertices B, C and D ofthe square ABCD. IfA(2, 1), C(2, 3) then focus ofthis parabola is - (A)       4 11 , 1 (B)       4 11 , 2 (C)       4 13 , 3 (D)       4 13 , 2 Solution : ClearlyAC is parallelto y-axis. It’s midpoint is (2, 2). Thus B  (1, 2). Parabola will be in the form of (x – 2)2 = (y– 3). It passes through (3, 2)  l = – . Thus parabola is (x – 2)2 = – 1(y – 3). It focus is x – 2 = 0. y – 3 = – 4 1 , i.e.,       4 11 , 2 .