Question

Find the point on the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=-1$ with eccentric angle $\frac{\pi }{6}$ radians.

• A. $(\frac{5}{\sqrt{3}},8\sqrt{3})$
• B. $(\frac{5}{\sqrt{3}},\frac{8}{\sqrt{3}})$
• C. $(\frac{8}{\sqrt{3}},\frac{5}{\sqrt{3}})$
• D. $(5\sqrt{3},\frac{8}{\sqrt{3}})$

Answer 1

The correct option is B $(\frac{5}{\sqrt{3}},\frac{8}{\sqrt{3}})$

Given: Hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=-1$, eccentric angle is $\frac{\pi }{6}$

To Find: Coordinates of the point with eccentric angle $\frac{\pi }{6}.$

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, $a=5,b=4.$

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 $P=(5\tan \frac{\pi }{6},4\sec \frac{\pi }{6})$

$\Rightarrow (\frac{5}{\sqrt{3}},\frac{8}{\sqrt{3}})$