Question

Tangents are drawn to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ at the ends of a latus rectum. The area of the quadrilateral to formed is A. Then (A/9) is

Answer 1

Ellipse: $\frac{x^2}{9}+\frac{y^2}{5}=1(e=\frac{2}{3})$

end of L.R$L:(2,\frac{5}{3})\&L^{\prime }(2,\frac{-5}{3})$

$T$ be the tangent at $L$ $\Rightarrow \frac{2x}{9}+\frac{y}{3}=1---\left(i\right)$

As in the intercept form so, intercept of $i$ on $y-$axis is $A(0,3)$.

$T^{\prime }$ be the tangent at $L^{\prime }=\frac{2x}{9}+\frac{-y}{3}=1---\left(ii\right)$

intercept of $ii$ on $y-$ axis is $B(0,-3)$

So the area of quadrilateral, $AB{L}^{\prime }L=\frac{1}{2}\left|\begin{array}{cc}0& 3\\ 2& \frac{5}{3}\\ 2& \frac{-5}{3}\\ 0& -3\\ 0& 3\end{array}\right|$

$A=\left|\frac{1}{2}\right(0-6-\frac{10}{3}-\frac{10}{3}-6-0\left)\right|$

$A=\frac{1}{2}\left(\frac{45}{3}\right)$

$A=23$

so, $\frac{A}{9}=\frac{23}{9}$