Equation of Normal to Hyperbola in all Forms

Equation of Normal to hyperbola : $$x^2\over a^2$$ – $$y^2\over b^2$$ = 1

(a) Point form :

The equation of normal to the given hyperbola at the point P($$x_1, y_1$$) is

$$a^2x\over x_1$$ + $$b^2y\over y_1$$ = $$a^2+b^2$$ = $$a^2e^2$$

Example : Find the equation of normal to the hyperbola $$x^2\over 25$$ – $$y^2\over 16$$ = 1 at (5, 1).

Solution : We have, $$x^2\over 25$$ – $$y^2\over 16$$ = 1

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

a = 5 and b = 4

Hence, required equation of normal is $$25x\over 5$$ + $$16y\over 1$$ = 41

= 5x + 16y = 41

(b) Slope form :

The normal to the given hyperbola whose slope is ‘m’, is

y = mx $$\mp$$ $${m(a^2+b^2)}\over \sqrt{a^2 – m^2b^2}$$

Foot of normal are ($$\mp$$$${a^2}\over \sqrt{a^2 – m^2b^2}$$, $$\mp$$$${mb^2}\over \sqrt{a^2 – m^2b^2}$$).

Example : 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

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

(c) Parametric form :

The equation of normal to the given hyperbola at its point (asec$$\theta$$, btan$$\theta$$), is

$$ax\over sec\theta$$ + $$by\over tan\theta$$ = $$a^2+b^2$$ = $$a^2e^2$$

Example : Line $$xcos\alpha$$ + $$ysin\alpha$$ = p is a normal to the hyperbola $$x^2\over a^2$$ – $$y^2\over b^2$$ = 1, if

Solution : We have, $$x^2\over a^2$$ – $$y^2\over b^2$$ = 1

The normal to hyperbola is $$ax\over sec\theta$$ + $$by\over tan\theta$$ = $$a^2+b^2$$

comparing it with the given line equation

$$acos\theta\over cos\alpha$$ = $$bcot\theta\over sin\alpha$$ = $$a^2 + b^2\over p$$

$$\implies$$ $$sec\theta$$ = $$ap\over cos\alpha(a^2+b^2)$$, $$tan\theta$$ = $$bp\over sin\alpha(a^2+b^2)$$

Eliminating $$\theta$$, we get

$$a^2sec^2\alpha$$ – $$b^2cosec^2\alpha$$ = $$(a^2 + b^2)^2)\over p^2$$