Question

If the radius of the auxiliary circle of hyperbola $\frac{x^2}{16}-\frac{y^2}{4}=1$ is $r$ and if the abscissa of the points with eccentric angle 60 degrees on the hyperbola and auxiliary circle respectively are $p$ and $q$, then what is the value of $p+q-r$ ?

Answer 1

Equation of auxiliary circle of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2$.

In this case, $a^2=16$.

Therefore, Required equation is $x^2+y^2=16$.

Corresponding points $N(4\cos 60,4\sin 60)$, $N(2,2\sqrt{3})$ on circle and

$P(4\sec 60,2\tan 60)$ , $P(8,2\sqrt{3})$ on hyperbola.

Hence $r=4,p=8,q=2$. So, $p+q-r=6$.