Question

Find the point on the hyperbola $\frac{(x+2{)}^2}{100}-\frac{(y-3{)}^2}{1}=1$ with eccentric angle $\frac{\pi }{4}$ radians.

• A. $(4,-2+10\sqrt{2})$
• B. $(-2+10\sqrt{2},4\sqrt{2})$
• C. $(-2+10\sqrt{2},4)$
• D. $(10\sqrt{2},4)$

Answer 1

The correct option is C $(-2+10\sqrt{2},4)$

Given: Hyperbola $\frac{(x+2{)}^2}{100}-\frac{(y-3{)}^2}{1}=1$, eccentric angle is $\frac{\pi }{4}$

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

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, $a=10,b=1,h=-2,k=3.$

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

Substituting the values of $a,b,h,k$ and $\theta$ we get,
Coordinates of $P(-2+\left(10\right)\sec \frac{\pi }{4},3+\left(1\right)\tan \frac{\pi }{4})$

$\Rightarrow P(-2+10\sqrt{2},3+1)$

$\Rightarrow P(-2+10\sqrt{2},3+1)$