Question

The parametric coordinates of the point $(8,3\sqrt{3})$ on the hyperbola $9x^2-16y^2=144$ is

• A. $(2\sec {60}^{\circ },3\tan {60}^{\circ })$
• B. $(3\sec {60}^{\circ },3\tan {60}^{\circ })$
• C. $(4\sec {60}^{\circ },3\tan {60}^{\circ })$
• D. $(4\sec {30}^{\circ },3\tan {30}^{\circ })$

Answer 1

The correct option is C $(4\sec {60}^{\circ },3\tan {60}^{\circ })$

We know that for all the values of $\theta$, the point $(a\sec \theta ,b\tan \theta )$ always lies on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Here $9x^2-16y^2=144$
$\Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1$
So, $(4\sec \theta ,3\tan \theta )=(8,3\sqrt{3})$
$\Rightarrow \sec \theta =2$ and $\tan \theta =\sqrt{3}$
$\Rightarrow \theta ={60}^{\circ }$
Thus coordinates are $(4\sec {60}^{\circ },3\tan {60}^{\circ })$