Question

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

Answer 1

Given: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Slope of tangent is$\frac{dy}{dx}:$
Differentiating w.r.t. $x,$
$\frac{2x}{a^2}-\frac{2y}{b^2}.\frac{dy}{dx}=0$
$\Rightarrow -\frac{y}{b^2}.\frac{dy}{dx}=-\frac{x}{a^2}$
$\Rightarrow \frac{dy}{dx}=\frac{b^2}{a^2}\left(\frac{x}{y}\right)$
Slope of tangent at $(x_0,y_0)$ is ${\begin{array}{c}\frac{dy}{dx}\end{array}|}_{(x_0,y_0)}=\frac{b^2}{a^2}\left(\frac{x_0}{y_0}\right)$
We know that,
$\text{Slope of tangent}\times \text{Slope of normal}=-1$
Slope of normal $=-\frac{a^2}{b^2}\left(\frac{y_0}{x_0}\right)$
Equation of tangent at $(x_0,y_0)\&$ having slope $\frac{b^2}{a^2}\left(\frac{x_0}{y_0}\right)$
$(y-y_0)=\frac{b^2}{a^2}\left(\frac{x_0}{y_0}\right)(x-x_0)$
$\Rightarrow \frac{y_{0}y-y_0^2}{b^2}=\frac{(x_{0}x-x_0^2)}{a^2}$
$\Rightarrow \frac{x_{0}x}{a^2}-\frac{y_{0}y}{b^2}=\frac{x_0^2}{a^2}-\frac{y_0^2}{b^2}$
Since point $(x_0,y_0)$ lie on the curve.
$\therefore$ it will satisfy the equation of curve $\frac{x_0^2}{a^2}-\frac{y_0^2}{b^2}=1$
Now, the equation of tangent becomes
$\Rightarrow \frac{x_{0}x}{a^2}-\frac{y_{0}y}{b^2}=1$
Equation of normal at $(x_0,y_0)\&$ having slope$-\frac{a^2}{b^2}\left(\frac{y_0}{x_0}\right)$ is $(y-y_0)=-\frac{a^2}{b^2}\left(\frac{y_0}{x_0}\right)(x-x_0)$
$\Rightarrow \frac{(y-y_0)}{a^2{y}_0}=-\frac{(x-x_0)}{b^2{x}_0}$
$\Rightarrow \frac{(y-y_0)}{a^2{y}_0}+\frac{(x-x_0)}{b^2{x}_0}=0$
Hence, the equation of tangent is $\frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=1$ and equation of normal is $\frac{y-y_0}{a^2{y}_0}+\frac{x-a_0}{b^2{x}_0}=0$