# Find the normal to the hyperbola $$x^2\over 16$$ – $$y^2\over 9$$ = 1 whose slope is 1.

## Solution :

We have, $$x^2\over 16$$ – $$y^2\over 9$$ = 1

Compare given equation with $$x^2\over a^2$$ – $$y^2\over b^2$$ = 1

a = 4 and b = 3

Since the normal to the given hyperbola whose slope is ‘m’, is  y = mx $$\mp$$ $${m(a^2+b^2)}\over \sqrt{a^2 – m^2b^2}$$

Hence, required equation of normal is y = x $$\mp$$ $${25}\over \sqrt{7}$$.

### Similar Questions

Angle between asymptotes of hyperbola xy=8 is

Find the asymptotes of the hyperbola $$2x^2 + 5xy + 2y^2 + 4x + 5y$$ = 0. Find also the general equation of all the hyperbolas having the same set of asymptotes.

Find the equation of the tangent to the hyperbola $$x^2 – 4y^2$$ = 36 which is perpendicular to the line x – y + 4 = 0

The eccentricity of the conjugate hyperbola to the hyperbola $$x^2-3y^2$$ = 1 is

If the foci of a hyperbola are foci of the ellipse $$x^2\over 25$$ + $$y^2\over 9$$ = 1. If the eccentricity of the hyperbola be 2, then its equation is :