# Equation of Tangent to Hyperbola in all Forms

## Equation of Tangent to Hyperbola $$x^2\over a^2$$ – $$y^2\over b^2$$ = 1

### (a) Point form :

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

$$x{x_1}\over a^2$$ – $$y{y_1}\over b^2$$ = 1

Example : Find the equation of tangent to the hyperbola $$16x^2$$ – $$9y^2$$ = 144 at (5, 16/3).

Solution : We have, $$16x^2$$ – $$9y^2$$ = 144

$$\implies$$ $$x^2\over 9$$ – $$y^2\over 16$$ = 1

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

a = 3 and b = 16

Hence, required equation of tangent is $$5x\over 9$$ – $$16/3\over 16$$ = 1

= $$5x\over 9$$ – $$y\over 3$$ = 1

### (b) Slope form :

The equation of tangent to the given hyperbola whose slope is ‘m’, is

y = mx $$\pm$$ $$\sqrt{a^2m^2 – b^2}$$

The Point of contact are ($${\mp} a^2m\over \sqrt{a^2m^2 – b^2}$$, $${\mp} b^2\over \sqrt{a^2m^2 – b^2}$$)

Note that there are two parallel tangents having the same slope m.

Example : Find the tangent to the hyperbola $$x^2\over 25$$ – $$y^2\over 16$$ = 1 whose slope is 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 y = x $$\pm$$ $$\sqrt{9}$$

$$\implies$$ y = x $$\pm$$ 3

### (c) Parametric form :

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

$$xsec\theta\over a$$ – $$ytan\theta\over b$$ = 1

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

Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line x – y = 0

$$\therefore$$   m$$\times$$1 = -1 $$\implies$$ m = -1

Since $$x^2-4y^2$$ = 36 or $$x^2\over 36$$ – $$y^2\over 9$$ = 1

Comparing this with $$x^2\over a^2$$ – $$y^2\over b^2$$ = 1

$$\therefore$$   $$a^2$$ = 36 and $$b^2$$ = 9

So the equation of the tangent are y = -1x $$\pm$$ $$\sqrt{36\times {-1}^2 – 9}$$

$$\implies$$ y = x $$\pm$$ $$\sqrt{27}$$ $$\implies$$ x + y $$\pm$$ 3$$\sqrt{3}$$ = 0