JEE Main 2015 · 4 Apr · Shift 1 · Mathematics Q81

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MathematicsSingle correct+4 / -1Question 81 of 90

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $\dfrac{x^2}9+\dfrac{y^2}5=1$ , is :

\dfrac{27}4

\dfrac{27}2

Answer and worked solution

Official answer: D

$e=\frac23$ , so the focus is at $ae=2$ , and the latus-rectum end is $\left(2,\frac53\right)$ . The tangent $\frac{2x}9+\frac{y}3=1$ cuts the axes at $\frac92$ and 3.

By symmetry the quadrilateral is a rhombus of area $4\times\frac12\times\frac92\times3=27$ .

Topics:Ellipse Tangents and normals to conics

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JEE Main 2015 · 4 Apr · Shift 1, Mathematics question 81