Question

If a quadrilateral formed by four tangents to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is a square then the area of the square is equal to

Answer 1

$\frac{x^2}{9}+\frac{y^2}{4}=1$

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{9}}=\frac{\surd 5}{3}$

$(ae,0)$ & $(-ae,0)$

$(\surd 5,0)$ & $(-\surd 5,0)$

$\frac{x{x}_1}{9}+\frac{y{y}_1}{4}=1$

$\frac{\surd 5x_1}{9}+\frac{\surd 5+3}{3}\times \frac{y}{4}=1$

$\frac{\surd 5x_1}{9}+\frac{(\surd +3)}{12}y=1$

Area of quad $=\frac{1}{2}\times \frac{\surd 5}{9}\times \frac{(\surd +3)}{12}$

$=\frac{\surd 5(\surd 5+3)}{18\times 12}sq.unit$

Area of square $=\frac{5+3\surd 5}{18\times 12}\times \frac{5+3\surd 5}{18\times 12}$

$=\frac{(5+3\surd 5{)}^2}{(18\times 12{)}^2}$

$=\frac{(11.708{)}^2}{(216{)}^2}$

$\Rightarrow \frac{137.0773}{46656}$

$=0.002938042$

$=0.0029$

$=0.003sq.unit$