Question

What will be the parametric equation of normal on a rectangular hyperbola $x^2-y^2=8$ if the eccentric angle is $\frac{\pi }{4}$?

• A. $x+y\sqrt{2}=4$
• B. $x+y\sqrt{2}=8$
• C. $x-y\sqrt{2}=8$

Answer 1

The correct option is B $x+y\sqrt{2}=8$

Given: Rectangular hyperbola $x^2-y^2=8$; Eccentric angle is $\frac{\pi }{4}$

To Find: Parametric equation of normal

We know that the parametric coordinates of a rectangular hyperbola are given by $(a\sec \theta ,a\tan \theta )$.

From the given hyperbola equation, we have $a=2\sqrt{2}$

The parametric equation of normal at a point on rectangular hyperbola is given by $x\tan \theta +y\sec \theta =2a\sec \theta \tan \theta$.

Substituting the values of $a$ and $\theta$.

$\Rightarrow x\tan (\frac{\pi }{4})+y\sec (\frac{\pi }{4})=2a\sec (\frac{\pi }{4})\tan (\frac{\pi }{4})$

$\Rightarrow x+y\sqrt{2}=2\times 2\sqrt{2}\times \sqrt{2}$

$\Rightarrow x+y\sqrt{2}=8$