24) Find the equations of the tangent and normal to the hyperbola
$\frac{x^2 }{a^2} - \frac{y^2 }{b^2 }= 1$ at the point $(x_0 , y_0 )$

Answers (1)

Given equation is
$\frac{x^2 }{a^2} - \frac{y^2 }{b^2 }= 1 \Rightarrow y^2a^2 = x^2b^2 -a^2b^2$
Now ,we know that
slope of tangent = $2ya^2\frac{dy}{dx} = 2xb^2 \Rightarrow \frac{dy}{dx} = \frac{xb^2}{ya^2}$
at point $(x_0 , y_0 )$
$\frac{dy}{dx} = \frac{x_0b^2}{y_0a^2}$
equation of tangent at point $(x_0 , y_0 )$ with slope $\frac{xb^2}{ya^2}$
$y-y_1=m(x-x_1)\\ y-y_0=\frac{x_0b^2}{y_0a^2}(x-x_0)\\ yy_0a^2-y_0^2a^2 = xx_0b^2-x_0^2b^2\\ xx_0b^2 - yy_0a^2 = x_0^2b^2-y_0^2a^2$
Now, divide both sides by $a^2b^2$
$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = \left ( \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} \right )$
$=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left ( \because \frac{x_0^2}{a^2}-\frac{y_0^2}{b^2 } = 1\right )$
$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$
Hence, the equation of tangent is

$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$
We know that
$Slope \ of \ normal= \frac{-1}{slope \ of \ tangent } = -\frac{y_0a^2}{x_0b^2}$
equation of normal at the point $(x_0 , y_0 )$ with slope $-\frac{y_0a^2}{x_0b^2}$
$y-y_1=m(x-x_1)\\ y-y_0=-\frac{y_0a^2}{x_0b^2}(x-x_0)\\ \frac{y-y_0}{y_0a^2} + \frac{x-x_0}{x_0b^2} = 0$