(A) If point P is on the circle x^2 + y^2 = 5, then the equation of the chord of contact with respect to the parabola

Question

Match the items of Column I with those of Column II.

Column I	Column II
(A) If point P is on the circle x2 + y2 = 5, then the equation of the chord of contact with respect to the parabola y2 = 4x is y = 2 (x - 2). The coordinates of P are	(p) (9, −6)
(B) Tangents are drawn from the point (2, 3) to a parabola y2 = 4x.	(q) (1, 2)
(C) The common chord of the circle x2 + y2 = 5 and the parabola 6y - 5x2 + 7x passes through	(r) (−2, 1)
(D) Two points P(4, −4) and Q are on the parabola y2 4x such that the area of ΔPOQ (O is the vertex) is 6 sq. unit. Then, the coordinates of Q are	(s) (4, 4)
	(t) (2, 1)

Answer 1

(A)  Let P (x₁, y₁) be on the circle x² + y² = 5. Then

x₁² + y₁²  = 5  ....(1)

The equation of the chord of contact of P(x_1, y₁) with the parabola y² = 4x is

yy₁ - 2(x + x₁) = 0

⇒ 2x - y₁y + 2x₁ = 0

However,

2x - y - 4 = 0 ....(2)

is the chord of contact. Therefore, from Eqs. (1) and (2), we get

Answer: (A) → (r)

(B) Tangent to the parabola y² = 4x at(t², 2t)  is ty = x + t². This passes through the point (2, 3). So

 Therefore, the points of contact are (1, 2) and (4, 4).

 Answer: (B) → (q), (s)

(C)  Substituting

y = 5x² + 7x/6

 in the circle equation x² + y² = 5, we get

which clearly implies that x = 1 is a root. So

Therefore, the points of intersection are (1, 2) and ( 2, 1).

Answer: (C) → (q), (r)

(D)  Let Q = (t², 2t). Therefore

Case 1:  When t² + 2t + 3 = 0, we have

 (t - 1)(t + 3) = 0

⇒ t = 1,-3

So Q = (1,2),(9, -6).

Case 2: When t² + 2t + 3 = 0, it has no real roots. 

Answer: (D) → (p), (q)