Question

For hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$, and circle $x^2+y^2=49$, Let $P,Q,R,S$ be concyclic points P in 1st and Q in 2nd qudrant.

Find slope of $PQ$.

• A. $1$
• B. $0$
• C. $\infty$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $0$

The given hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and given circle $x^2+y^2=49$, cut each-other at four common points, which are con-cyclic points.

From equation of circle getting $y^2=49-x^2$ ....$\left(1\right)$

and putting value of $y^2$ from equation $\left(1\right)$ into equation of given hyperbola we get,

$\to \frac{x^2}{16}-\frac{(49-x^2)}{9}=1$

$\to \frac{x^2}{16}+\frac{x^2}{9}-\frac{\left(49\right)}{9}=1$

$\to \frac{25x^2}{144}=\frac{\left(58\right)}{9}$

$\to x=\pm \frac{4\sqrt{58}}{5}$

$\to x_P=+\frac{4\sqrt{58}}{5}$

$\to x_Q=-\frac{4\sqrt{58}}{5}$

Now $y^2=49-x^2=49-$ $(\pm \frac{4\sqrt{58}}{5}{)}^2$

$\to y^2=\left(\frac{1225-928}{25}\right)$

$\to y^2=\pm \frac{297}{25}$

$\to y=\pm \frac{\sqrt{297}}{5}$

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

Hence, $\to y_P=+\frac{\sqrt{297}}{5}$

$\to y_Q=+\frac{\sqrt{297}}{5}$

Slope of line $PQ,$ $m=\frac{y_Q-y_P}{x_Q-x_P}$

As $y_Q=y_P$, hence the slope of line $PQ$ is $0$

Correct option is $B$.