Question

Find the point on the hyperbola $\frac{x^2}{36}-\frac{y^2}{9}=1$ with eccentric angle $\frac{\pi }{4}$ radians.

Answer 1

Given: Hyperbola $\frac{x^2}{36}-\frac{y^2}{9}=1,$ eccentric angle is $\frac{\pi }{4}$

To Find: Coordinates of the point with eccentric angle $\frac{\pi }{4}.$

Step - $1$: Find the values of $a$ and $b$

Step - $2$: Substitute the values of $a$ and $b$ in the standard parametric coordinates of a point on the hyperbola.

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

On comparing the given hyperbola equation with the standard equation, $a=6,b=3.$

Any point $P\left(\theta \right)$ on the hyperbola: $(a\sec \theta ,b\tan \theta )$

Substituting the values of $a$ and $b$ we get,
Coordinates of $P=(6\sec \frac{\pi }{4},3\tan \frac{\pi }{4})$

$\Rightarrow (6\sqrt{2},3)$