Question

Find the point on the hyperbola $\frac{(x-1{)}^2}{25}-\frac{(y-1{)}^2}{9}=-1$ with eccentric angle $\frac{\pi }{6}$ radians.

• A. $(1+\frac{6}{\sqrt{3}},1+\frac{5}{\sqrt{3}})$
• B. $(-1+\frac{5}{\sqrt{3}},-1+\frac{6}{\sqrt{3}})$
• C. $(1+\frac{5}{\sqrt{2}},1+\frac{6}{\sqrt{2}})$
• D. $(1+\frac{5}{\sqrt{3}},1+\frac{6}{\sqrt{3}})$

Answer 1

The correct option is D $(1+\frac{5}{\sqrt{3}},1+\frac{6}{\sqrt{3}})$

Given: $\text{Hyperbola}\frac{(x-1{)}^2}{25}-\frac{(y-1{)}^2}{9}=-1$, 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 translated conjugate hyperbola.

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

On comparing the given hyperbola equation with the standard equation, $a=5,b=3,h=1,k=1$.

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

Substituting the values of $a,b,h,k$ and $\theta$ we get,
Coordinates of $P=(1+\left(5\right)\tan \frac{\pi }{6},1+3\sec \frac{\pi }{6})$

$\Rightarrow P=(1+5\frac{1}{\sqrt{3}},1+3\frac{2}{\sqrt{3}})$

$\Rightarrow P=(1+\frac{5}{\sqrt{3}},1+\frac{6}{\sqrt{3}})$