JEE Main 2022 · 28 Jul · Shift 2 · Mathematics Q83

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MathematicsInteger+4 / -1Question 83 of 90

Let the tangents at the points P and Q on the ellipse $\dfrac{x^2}2+\dfrac{y^2}4=1$ meet at the point $R\left(\sqrt2,2\sqrt2-2\right)$ . If S is the focus of the ellipse on its negative major axis, then $SP^2+SQ^2$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 13

Chord of contact from R: $\frac{\sqrt2x}2+\frac{(2\sqrt2-2)y}4=1$ . It meets the ellipse at $P(1,\sqrt2)$ and $Q(\sqrt2,0)$ .

Focus. The major axis is the y-axis ( $b^2=4$ ), with $e=\frac1{\sqrt2}$ , so $S=(0,-\sqrt2)$ .

Sum $SP^2+SQ^2=(1+8)+(2+2)=13$ .

Answer: 13

Topics:Tangents and normals to conics Ellipse

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JEE Main 2022 · 28 Jul · Shift 2, Mathematics question 83