Consider the ellipse 9x2+4y2=1. Let S(p,q) be a point in the first quadrant such that 9p2+4q2>1. Two tangents are drawn from S to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point T in the fourth quadrant. Let R be the vertex of the ellipse with positive x-coordinate and O be the center of the ellipse. If the area of the triangle Δ ORT is 23, then which of the following options is correct ?
(A)q=2,p=33✓ Correct
(B)q=2,p=43
(C)q=1,p=53
(D)q=1,p=63
Answer: (A)
Solution
The ellipse is 9x2+4y2=1 with a=3, b=2.
One tangent from S(p,q) touches at (0,2) (top of minor axis). The tangent at (0,2) is y=2.
The vertex R with positive x-coordinate is (3,0).
Area of △ORT=23 where O=(0,0), R=(3,0), and T is in the fourth quadrant.
Using Area=21∣xR⋅yT∣=21⋅3⋅∣yT∣=23, we get ∣yT∣=1, so yT=−1.
Substituting in the ellipse equation: 9xT2+41=1 gives xT2=427, so xT=233.
Thus T(233,−1).
The tangent at T is 9x⋅233+4y⋅(−1)=1, which simplifies to 63x−4y=1.
Since both tangents pass through S(p,q), substituting (p,q) in y=2 gives q=2.
Substituting in the second tangent: 63p−42=1 gives 63p=23, so p=33.
JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Single correct option
