Find the equations of the tangent and normal to the hyperbola

Solution:

Differentiating $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with respect to $x$, we have:

$\frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \frac{d y}{d x}=0$

$\Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x}=\frac{2 x}{a^{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{b^{2} x}{a^{2} y}$

Therefore, the slope of the tangent at $\left(x_{0}, y_{0}\right)$ is $\left.\frac{d y}{d x}\right]_{\left(x_{0}, y_{0}\right)}=\frac{b^{2} x_{0}}{a^{2} y_{0}}$.

Then, the equation of the tangent at $\left(x_{0}, y_{0}\right)$ is given by,

$y-y_{0}=\frac{b^{2} x_{0}}{a^{2} y_{0}}\left(x-x_{0}\right)$

$\Rightarrow a^{2} y y_{0}-a^{2} y_{0}^{2}=b^{2} x x_{0}-b^{2} x_{0}^{2}$

$\Rightarrow b^{2} x x_{0}-a^{2} y y_{0}-b^{2} x_{0}^{2}+a^{2} y_{0}^{2}=0$

$\Rightarrow \frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}-\left(\frac{x_{0}^{2}}{a^{2}}-\frac{y_{0}^{2}}{b^{2}}\right)=0 \quad\left[\right.$ On dividing both sides by $\left.a^{2} b^{2}\right]$

$\Rightarrow \frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}-1=0 \quad\left[\left(x_{0}, y_{0}\right)\right.$ lies on the hyperbola $\left.\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\right]$

$\Rightarrow \frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}=1$

Now, the slope of the normal at $\left(x_{0}, y_{0}\right)$ is given by,

$\frac{-1}{\text { Slope of the tangent at }\left(x_{0}, y_{0}\right)}=\frac{-a^{2} y_{0}}{b^{2} x_{0}}$

Hence, the equation of the normal at $\left(x_{0}, y_{0}\right)$ is given by,

$y-y_{0}=\frac{-a^{2} y_{0}}{b^{2} x_{0}}\left(x-x_{0}\right)$

$\Rightarrow \frac{y-y_{0}}{a^{2} y_{0}}=\frac{-\left(x-x_{0}\right)}{b^{2} x_{0}}$

$\Rightarrow \frac{y-y_{0}}{a^{2} y_{0}}+\frac{\left(x-x_{0}\right)}{b^{2} x_{0}}=0$