Question

From the point $P(1,5)$ tangents $PQ$ and $PR$ are drawn to the ellipse $\frac{x^2}{4}+\frac{y^2}{25}=1$, then the acute angle $QOR,$ where $O$ is the origin, is

• A. ${\tan }^{-1}\frac{8}{15}$
• B. ${\tan }^{-1}\frac{8}{35}$
• C. ${\tan }^{-1}\frac{15}{8}$
• D. ${\tan }^{-1}\frac{4}{15}$

Answer 1

The correct option is A ${\tan }^{-1}\frac{8}{15}$

Equation of chord of contact $QR$ is
$T=0$
$\frac{x}{4}+\frac{5y}{25}=1$
$\Rightarrow \frac{x}{4}+\frac{y}{5}=1\cdots \left(1\right)$

Homogenising the given equation we will get $OQ$ and $OR$
Using Equation $\left(1\right)$
$\frac{x^2}{4}+\frac{y^2}{25}={(\frac{x}{4}+\frac{y}{5})}^2$
$\Rightarrow \frac{x^2}{4}+\frac{y^2}{25}=\frac{x^2}{16}+\frac{y^2}{25}+\frac{2xy}{20}\Rightarrow \frac{3x^2}{16}=\frac{2xy}{20}\Rightarrow 15x^2-8xy=0$

So,
$\tan \theta =|\frac{\sqrt{h^2-ab}}{a+b}|\Rightarrow \tan \theta =|\frac{\sqrt{8^2}}{15}|=\frac{8}{15}$
$\Rightarrow \theta ={\tan }^{-1}\frac{8}{15}$