Let the tangent and normal at the point on the ellipse meet the -axis at the points and , respectively. Let the circle be drawn taking as a diameter and the line intersect at the points and . If the tangents at the points and on the circle intersect at the point , then is equal to:

Explanation: 1. Tangent aur Normal ki Equations ( -axis par points ) Ellipse ke liye point par: Tangent ki equation: -axis ke liye rakhein: . Toh point . Normal ki equation: -axis ke liye rakhein: . Toh point . 2. Circle ki Equation aur diameter ke endpoints hain. Center = Radius = Equation: 3. Intersection Point Line circle ki Chord of Contact hai point ke liye. Circle ke liye point se Chord of Contact ki equation: Hume pata hai ye line (ya ) hi hai. Dono ko compare karne par: ka coefficient hona chahiye: Constants aur ke ratio: ki value rakhne par: 4. Final Calculation ( )

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let the tangent and normal at the point $(3\sqrt{3}, 1)$ on the ellipse $\frac{x^2}{36} + \frac{y^2}{4} = 1$ meet the $y$ -axis at the points $A$ and $B$ , respectively. Let the circle $C$ be drawn taking $AB$ as a diameter and the line $x = 2\sqrt{5}$ intersect $C$ at the points $P$ and $Q$ . If the tangents at the points $P$ and $Q$ on the circle intersect at the point $(\alpha, \beta)$ , then $\alpha^2 - \beta^2$ is equal to:

JEE 2023 — Mathematics PYQ

Choose the correct answer:

Explanation

1. Tangent aur Normal ki Equations ( $y$ -axis par points $A, B$ )

2. Circle $C$ ki Equation

3. Intersection Point $(\alpha, \beta)$

4. Final Calculation ( $\alpha^2 - \beta^2$ )

Explanation

1. Tangent aur Normal ki Equations ( $y$ -axis par points $A, B$ )

2. Circle $C$ ki Equation

3. Intersection Point $(\alpha, \beta)$

4. Final Calculation ( $\alpha^2 - \beta^2$ )