Question

If the director circle of ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ of eccentricity $\frac{1}{\sqrt{3}}$ passes through the two foci of the hyperbola $H : \frac{ y ^2}{9}-\frac{ x ^2}{16}=1$ then

• A. equation of director circle of ellipse is $x^2+y^2=25$.
• B. Area of quadrilateral formed by tangents at ends of latus rectum of ellipse $E$ is $30 \sqrt{3}$ sq. units.
• C. Area of square formed by four tangents to ellipse $E$ is 50 sq. units
• D. number of points on ellipse from which two perpendicular tangents can be drawn to hyperbola $H$ is 4 .

Answer 1

Answer: (A), (B), (C)

$e _{ H }^2=1+\frac{16}{9}=\frac{25}{9} \Rightarrow e _{ H }=\frac{5}{3} \Rightarrow \text { foci }(0, \pm 5)$
$\therefore a ^2+ b ^2=25 \text { and } a ^2- b ^2=\frac{ a ^2}{3} \Rightarrow \frac{2 a ^2}{3}= b ^2 $
$\frac{5 a ^2}{3}=25 \Rightarrow a ^2=15 \text { and } b ^2=10$
$\therefore E : \frac{ x ^2}{15}+\frac{ y ^2}{10}=1 \Rightarrow A =\frac{2 a ^2}{ e }=\frac{2 \cdot 15}{\frac{1}{\sqrt{3}}}=30 \sqrt{3}$
Area of square formed by tangents $=\left(\frac{2 R_{\text {Director }}}{\sqrt{2}}\right)^2=\frac{4 \cdot 25}{2}=50$ sq. units