Question

For hyperbola $\frac{x^2}{25}-\frac{y^2}{12}=1$, and circle $x^2+y^2=100$, Let P,Q,R,S be concyclic points.Angle between PR and QS is

• A. ${\tan }^{-1}\left(\frac{60}{20\sqrt{7}+30}\right)$
• B. ${\tan }^{-1}\left(\frac{60}{20\sqrt{7}-30}\right)$
• C. $\frac{\pi }{2}$
  
• D. None of the above

Answer 1

The correct option is D None of the above

The equation of hyperbola is $\frac{x^2}{25}-\frac{y^2}{12}=0$

The equation of circle is $x^2+y^2=100$

Now substitute $x^2=\frac{25}{12}{y}^2$ in the circle equation

We get $\frac{37}{12}{y}^2=100$

$\Rightarrow y=\pm \frac{20\sqrt{3}}{\sqrt{37}}$

Therefore $x=\pm \frac{50}{\sqrt{37}}$

The four con-cyclic points are $P=(\frac{50}{\sqrt{37}},\frac{20\sqrt{3}}{\sqrt{37}})$ , $Q=(-\frac{50}{\sqrt{37}},\frac{20\sqrt{3}}{\sqrt{37}})$ , $R=(-\frac{50}{\sqrt{37}},-\frac{20\sqrt{3}}{\sqrt{37}})$ , $S=(\frac{50}{\sqrt{37}},-\frac{20\sqrt{3}}{\sqrt{37}})$

Slope of $PR$ is $\frac{2\sqrt{3}}{5}$ and the slope of $QS$ is $-\frac{2\sqrt{3}}{5}$

Let the angle between them be $\theta$

We have $tan\left(\theta \right)=|\frac{\frac{2\sqrt{3}}{5}+\frac{2\sqrt{3}}{5}}{1-\frac{2\sqrt{3}}{5}\times \frac{2\sqrt{3}}{5}}|=\frac{20\sqrt{3}}{13}$