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Common Chord of Circle and y²=4ax Divides VF in Ratio | JEE

JEE Maths question with a full step-by-step solution.

Question
A circle is described whose centre and diameter are the vertex and three quarters of the latus rectum of the parabola y2=4axy^2 = 4ax, respectively. The common chord of the circle and the parabola divides the distance between the vertex and the focus in the ratio of:
A1:11:1correct
B1:21:2
C1:31:3
D3:13:1
Solution
Step 1: Centre OO, radius 12⋅34(4a)=3a2\dfrac{1}{2}\cdot\dfrac{3}{4}(4a) = \dfrac{3a}{2}, so x2+y2=9a24x^2 + y^2 = \dfrac{9a^2}{4}. Eliminating y2y^2,
x2+4ax−9a24=0⇒(2x−a)(2x+9a)=0⇒x=a2  (x≥0).x^2 + 4ax - \frac{9a^2}{4} = 0 \Rightarrow (2x - a)(2x + 9a) = 0 \Rightarrow x = \frac{a}{2}\ \ (x \ge 0).
 Step 2: The vertex is (0,0)(0,0) and the focus (a,0)(a, 0); the chord meets the axis at x=a2x = \dfrac{a}{2}, the midpoint, so the ratio is 1:11 : 1. Correct answer: (1)
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