Question

For hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$, and circle $x^2+y^2=100$, let $P$ and $Q$ be concyclic points in first and second quadrant respectively, find $l\left(PQ\right)$.

• A. $\frac{10\sqrt{29}}{\sqrt{41}}$
• B. $\frac{20\sqrt{29}}{\sqrt{41}}$
• C. $\frac{30\sqrt{29}}{\sqrt{41}}$
• D. $\frac{60\sqrt{29}}{\sqrt{41}}$

Answer 1

The correct option is B $\frac{20\sqrt{29}}{\sqrt{41}}$

The given hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ and given circle $x^2+y^2=100$ cut each-other at four common points, which are con-cyclic points.

From equation of circle $y^2=100-x^2$

By putting value of $y^2$ from equation of circle into equation of hyperbola we get,

$\frac{x^2}{25}-\frac{(100-x^2)}{16}=1$

$\frac{x^2}{25}+\frac{x^2}{16}-\frac{\left(100\right)}{16}=1$

$\frac{41x^2}{25\times 16}=\frac{\left(29\right)}{4}$

$x=\pm \frac{10\sqrt{29}}{\sqrt{41}}$

$x_P=+\frac{10\sqrt{29}}{\sqrt{41}}$

$x_Q=-\frac{10\sqrt{29}}{\sqrt{41}}$

Now $y^2=100-x^2=100-$ $(\pm \frac{10\sqrt{29}}{\sqrt{41}}{)}^2$

$y^2=100\left(\frac{41-29}{41}\right)$

$y=\pm \frac{20\sqrt{3}}{41}$

As in both quadrants, $1st$ and $2nd$, the value of $y$ coordinate is a positive value.

So, $y_P=+\frac{20\sqrt{3}}{41}$

$y_Q=+\frac{20\sqrt{3}}{41}$

As $y$ coordinate for points $p$ and $Q$ is same, hence length of $PQ$ or $l_{PQ}=x_p-x_Q$

$l_{PQ}=\frac{20\sqrt{29}}{41}$