# find the equation of the tangent and normal to the hyperbole s2/a2 - y2/b2 = 1 at the point (x0,y0)

Given eqn. is $\large\frac{ x^2}{a^2} - \frac{y^2}{b^2}$$= 1.....(i) Differentiating on both sides w.r.t x we get, \large\frac{2x}{a^2} -\frac{2y}{b^2}.\frac{dy}{dx}$$ = 0$

$\Rightarrow\:\large\frac{ dy}{dx} =\frac{ b^2}{a^2}.\frac{x}{y}$

$\large\frac{dy}{dx }$  at  $( x_0,y_0)$ is $\large\frac{b^2}{a^2}.\frac{x_0}{y_0}$

$\therefore\:$  slope of  tangent  $m=\large\frac{b^2}{a^2}.\frac{x_0}{y_0}$

The equation of the tangent  at $(x_0,y_0)$  is

$y-y_0 = m(x-x_0)$   where  $m$  is slope of tangent

$\Rightarrow\: y-y_0 = \large\frac{b^2}{a^2}.\frac{x_0}{y_0}$$.(x-x_0) on simplifying we get, xx_0b^2 - yy_0a^2 = x_0^2.b^2 - y_0^2.a^2 dividing throughout by a^2.b^2 we get \large\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = \frac{x_0^2}{a^2 }- \frac{y_0^2}{b^2} that is the eqn. of tangent is \large\frac{xx_0}{a^2} - \frac{yy_0}{b^2 }$$= 1$  (from  (i)).

Equation of the normal is

$y-y_0=\large\frac{-1}{m}$$.(x-x_0) i.e., y-y_0=\large\frac{ -a^2y_0}{b^2x_0}$$(x-x_0)$

$\Rightarrow\: b^2x_0y - b^2x_0y_0 = - a^2y_0x +a^2y_0x_0$

that is $b^2x_0y + a^2y_0x = b^2x_0y_0 + a^2x_0y_0$

Dividing both the sides by $a^2b^2$  we get

The eqn. of the normal as

$\large\frac{x_0}{a^2}.$$y +\large\frac{y_0}{b^2}.$$x = \large\frac{x_0y_0}{a^2}$$+\large\frac{x_0y_0}{b^2}$
edited Oct 26, 2013