Question 1

Let O be the vertex of the parabola y^2=4x and its chords OP and OQ are perpendicular to each other. If the locus of the mid-point of the line segment PQ is a conic C, then the length of its latus rec

Accepted Answer

Let the vertex of the parabola y^2 = 4x be O(0, 0). Let the coordinates of the points P and Q on the parabola be (t_1^2, 2t_1) and (t_2^2, 2t_2) respectively. The slope of the chord OP is m_1 = 2t_1 - 0 t_1^2 - 0 = 2 t_1 . The slope of the chord OQ is m_2 = 2t_2 - 0 t_2^2 - 0 = 2 t_2 . Since OP and OQ are perpendicular to each other, m_1 m_2 = -1. ( 2 t_1 ) ( 2 t_2 ) = -1 t_1 t_2 = -4 Let M(h, k) be the mid-point of the line segment PQ. Then, h = t_1^2 + t_2^2 2 t_1^2 + t_2^2 = 2h k = 2t_1 + 2t_

Question 2

Let chord PQ of length 3 13 of the parabola y^2 = 12x be such that the ordinates of points P and Q are in the ratio 1:2. If the chord PQ subtends an angle at the focus of the parabola, then is equal t

Accepted Answer

Let the coordinates of points P and Q on the parabola y^2 = 12x be (x_1, y_1) and (x_2, y_2). Given that the ordinates are in the ratio 1:2, we have y_2 = 2y_1. Since P and Q lie on the parabola, their abscissae are x_1 = y_1^2 12 and x_2 = y_2^2 12 = 4y_1^2 12 = y_1^2 3 . The length of the chord PQ is 3 13 , so PQ^2 = 117. Using the distance formula: (x_2 - x_1)^2 + (y_2 - y_1)^2 = 117 ( y_1^2 3 - y_1^2 12 )^2 + (2y_1 - y_1)^2 = 117 ( y_1^2 4 )^2 + y_1^2 = 117 y_1^4 16 + y_1^2 - 117 = 0 y_1^4 +

Question 3

Let the point P be the vertex of the parabola y = x^2 - 6x + 12. If a line passing through the point P intersects the circle x^2 + y^2 - 2x - 4y + 3 = 0 at the points R and S, then the maximum value o

Accepted Answer

The equation of the parabola is y = x^2 - 6x + 12, which can be rewritten as y - 3 = (x - 3)^2. The vertex of the parabola is P(3, 3). The equation of the circle is x^2 + y^2 - 2x - 4y + 3 = 0, which can be rewritten as (x - 1)^2 + (y - 2)^2 = 2. The center of the circle is C(1, 2) and its radius is r = 2 . The distance between the point P and the center C is: PC = (3 - 1)^2 + (3 - 2)^2 = 4 + 1 = 5 Since PC = 5 > r = 2 , the point P lies outside the circle. Let the line through P intersect the c

Question 4

Let the directrix of the parabola P: y^2 = 8x, cut x-axis at the point A. Let B( , ), > 1, be a point on P such that the slope of AB is 3/5. If BC is a focal chord of P, then six times the area of ABC

Accepted Answer

For the parabola P: y^2 = 8x, we have 4a = 8 a = 2. The equation of the directrix is x = -a x = -2. Since the directrix cuts the x-axis at A, the coordinates of A are (-2, 0). Let the coordinates of point B on the parabola be (2t_1^2, 4t_1). The slope of AB is given as 3 5 . 4t_1 - 0 2t_1^2 - (-2) = 3 5 2t_1 t_1^2 + 1 = 3 5 10t_1 = 3t_1^2 + 3 3t_1^2 - 10t_1 + 3 = 0 (3t_1 - 1)(t_1 - 3) = 0 t_1 = 1 3 or t_1 = 3. For t_1 = 1 3 , = 2 ( 1 3 )^2 = 2 9 , which is rejected since > 1. For t_1 = 3, = 2(3)

Question 5

Let A,B and C be the vertices of a variable right angled triangle inscribed in the parabola y^2=16x. Let the vertex B containing the right angle be (4,8) and the locus of the centroid of ABC be a coni

Accepted Answer

The equation of the parabola is y^2 = 16x, which gives a = 4. Any point on the parabola can be taken as (4t^2, 8t). The vertex B is given as (4,8), which corresponds to the parameter t = 1. Let the parameters for vertices A and C be t_1 and t_2 respectively. Thus, A (4t_1^2, 8t_1) and C (4t_2^2, 8t_2). The slope of AB is: m_ AB = 8t_1 - 8 4t_1^2 - 4 = 2(t_1 - 1) (t_1 - 1)(t_1 + 1) = 2 t_1 + 1 Similarly, the slope of BC is: m_ BC = 2 t_2 + 1 Since ABC is right-angled at B, AB BC. Therefore, m_ AB

Question 6

Consider the parabola P : y^2 = 4kx and the ellipse E : x^2 a^2 + y^2 b^2 = 1. Let the line segment joining the points of intersection of P and E, be their latus rectums. If the eccentricity of E is e

Accepted Answer

The latus rectum of the parabola P: y^2 = 4kx is the line segment x = k. The endpoints of this latus rectum are (k, 2k) and (k, -2k). The latus rectum of the ellipse E: x^2 a^2 + y^2 b^2 = 1 is the line segment x = ae (taking the one in the positive x-axis). The endpoints of this latus rectum are (ae, b^2 a ) and (ae, - b^2 a ). Since the line segment joining the points of intersection of P and E is the latus rectum for both curves, their endpoints must coincide. Comparing the coordinates, we ge

Question 7

Let the parabola y = x^2 + px + q passing through the point (1, -1) be such that the distance between its vertex and the x-axis is minimum. Then the value of p^2 + q^2 is:

Accepted Answer

Given the equation of the parabola y = x^2 + px + q. Since it passes through (1, -1), we have: -1 = 1 + p + q q = -p - 2 The vertex of the parabola y = x^2 + px + q occurs at x = - p 2 . The y-coordinate of the vertex is: y_v = (- p 2 )^2 + p (- p 2 ) + q = - p^2 4 + q The distance between the vertex and the x-axis is |y_v|. Substituting q = -p - 2 into the expression for y_v: |y_v| = |- p^2 4 - p - 2 | = 1 4 |p^2 + 4p + 8| = 1 4 |(p+2)^2 + 4| To minimize this distance, the term (p+2)^2 must be

Question 8

Let A be the focus of the parabola y^ 2 =8 x. Let the line y= m x+ c intersect the parabola at two distinct points B and C. If the centroid of the triangle A B C is ( 7 3 , 4 3 ), then (B C)^ 2 is equ

Accepted Answer

Coordinates of centroid of triangle ABC are 2 3 (t_1^2 + t_2^2 + 1) = 7 3 t_1^2 + t_2^2 = 5 2 4 3 (t_1 + t_2) = 4 3 t_1 + t_2 = 1 (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1 t_2 t_1 t_2 = -3 4 (t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1 t_2 = 4 (BC)^2 = 4(t_1^2 - t_2^2)^2 + 16(t_1 - t_2)^2 (BC)^2 = 80

Question 9

Let the image of parabola x^ 2 =4 y, in the line x-y=1 be (y+a)^ 2 =b(x-c), a, b, c ~N . Then a+b+c is equal to

Accepted Answer

Reflection of point (h, k) in line x - y - 1 = 0: x-h 1 = y-k -1 = -2(h-k-1) 2 = k - h + 1. Image: (k+1,\ h-1). Let (X, Y) = (k+1,\ h-1), so h = Y+1,\ k = X-1. From h^2 = 4k: (Y+1)^2 = 4(X-1). Comparing with (y+a)^2 = b(x-c): a = 1,\ b = 4,\ c = 1. a + b + c = 1 + 4 + 1 = 6.

Question 10

An equilateral triangle OAB is inscribed in the parabola y^ 2 =4 x with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having A B as a diameter from the origin is

Accepted Answer

Parabola y^2 = 4x has vertex O = (0,0). By symmetry, let A = (t^2, 2t), B = (t^2, -2t). OA = t t^2+4 , AB = 4t. For equilateral: OA = AB t t^2+4 = 4t t^2 = 12 t = 2 3 . A = (12, 4 3 ), B = (12, -4 3 ). Circle with AB as diameter: centre (12, 0), radius = 4 3 . Minimum distance from origin = 12 - 4 3 = 4(3 - 3 ).

Question 11

Let the locus of the mid-point of the chord through the origin O of the parabola y^ 2 =4 x be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the rati

Accepted Answer

y^2 = 4x Locus of mid point of OP M(h, k) h = t^2 2 , k = t k^2 = 2h y^2 = 2x S: y^2 = 2x R(h, k) h = 3t^2 4 , k = 3t 4 t^2 = 8h 3 , t = 4k 3 16k^2 9 = 8h 3 2k^2 = 3h Locus of R : 2y^2 = 3x

Question 12

If the chord joining the points P _ 1 (x_ 1 , y_ 1 ) and P _ 2 (x_ 2 , y_ 2 ) on the parabola y^ 2 =12 x subtends a right angle at the vertex of the parabola, then x_ 1 x_ 2 -y_ 1 y_ 2 is equal to

Accepted Answer

Parabola y^2 = 12x has 4a = 12, so a = 3. Parametric points: P_1 = (3t_1^2, 6t_1), P_2 = (3t_2^2, 6t_2). Chord subtends right angle at vertex O(0,0): OP_1 OP_2 = 0. 9t_1^2 t_2^2 + 36t_1 t_2 = 0 t_1 t_2(t_1 t_2 + 4) = 0 t_1 t_2 = -4. x_1 x_2 - y_1 y_2 = 9(t_1 t_2)^2 - 36(t_1 t_2) = 9(16) - 36(-4) = 144 + 144 = 288.

Question 13

Let y^ 2 =12 x be the parabola with its vertex at O. Let P be a point on the parabola and A be a point on the x-axis such that OPA =90^ . Then the locus of the centroid of such triangles OPA is :

Accepted Answer

Parabola y^2 = 12x, a = 3. Let P = (3t^2, 6t), O = (0,0), A = (h, 0). OPA = 90^ PO PA = 0: (-3t^2)(h - 3t^2) + (-6t)(-6t) = 0 h = 3t^2 + 12. Centroid G = ( 0 + 3t^2 + h 3 , 0 + 6t + 0 3 ) = (2t^2 + 4, \ 2t). Let G = (X, Y): t = Y/2, X = Y^2/2 + 4. Locus: y^2 - 2x + 8 = 0.

Question 14

Let one end of a focal chord of the parabola y^ 2 =16 x be (16,16). If P ( , ) divides this focal chord internally in the ratio 5: 2, then the minimum value of + is equal to :

Accepted Answer

Parabola y^2 = 16x, a = 4, focus at (4, 0). Point (16, 16) has parameter t_1 = 2. Other end of focal chord: t_2 = -1/t_1 = -1/2, giving (4 1/4, 8 (-1/2)) = (1, -4). P divides the chord in ratio 5:2 internally (two cases): From (16,16) to (1,-4): P = ( 37 7 , 12 7 ), + = 7. From (1,-4) to (16,16): P = ( 82 7 , 72 7 ), + = 22. Minimum value of + = 7.

Question 15

Let O be the vertex of the parabola x^ 2 =4 y and Q be any point on it. Let the locus of the point P, which divides the line segment OQ internally in the ratio 2: 3 be the conic C. Then the equation o

Accepted Answer

Parabola x^2 = 4y, vertex O = (0, 0), Q = (2t, t^2) on parabola. P divides OQ in ratio 2:3: P = ( 4t 5 , 2t^2 5 ). Eliminating t: x^2 = 8y 5 (conic C with a = 2 5 ). For chord with midpoint (h, k): hx - 2ay = h^2 - 2ak. At (1, 2): x - 4y 5 = 1 - 8 5 = - 3 5 . 5x - 4y + 3 = 0.

Parabola — Previous Year Questions & Solutions