The area of the quadrilateral formed by the tangents at the end point of latus rectum to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ is $3 K$ . Then $K$ is equal to
$∴ area = 4 [\frac{1}{2} \cdot \frac{9}{2} \cdot 3] = 27$

Question

The area of the quadrilateral formed by the tangents at the end point of latus rectum to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ is $3 K$ . Then $K$ is equal to

$∴ area = 4 [\frac{1}{2} \cdot \frac{9}{2} \cdot 3] = 27$

TestHub · Accepted Answer

e = 2/3 Equation of tangent at L is 9 2 x ​ + 3 y ​ = 1 it meets x -axis at A ( 2 9 ​ , 0 ) \& y axis at B ( 0 , 3 ) .

Mathematics - Ellipse Question with Solution | TestHub

Doubts & Discussion