Question

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 $(x_0,y_0)$.

Answer 1

The equation of the given curve is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .....(i)

On differentiating both sides w.r.t x, we get

$\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$

$\therefore$ The slope of the tangent at $(x_0,y_0)$ =${\left(\frac{dy}{dx}\right)}_{(x_0,y_0)}=\frac{b^2{x}_0}{a^2{y}_0}$

Hence, the equation of the tangent at $(x_0,y_0)$ is given by

$y-y_0=\frac{b^2{x}_0}{a^2{y}_0}(x-x_0)\Rightarrow \frac{y_0}{b^2}(y-y_0)=\frac{x_0}{a^2}(x-x_0)$

$\Rightarrow \frac{y{y}_0}{b^2}-\frac{y_0^2}{b^2}=\frac{x{x}_0}{a^2}-\frac{x_0^2}{a^2}\Rightarrow \frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=\frac{x_0^2}{b^2}-\frac{y_0^2}{b^2}\Rightarrow \frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=1$

[Using the fact that $(x_0,y_0)$ lies on Eq. (i) as a hyperbola $\frac{x^2}{a^a}-\frac{y^2}{b^a}=1$]

Now, the equation of the normal at $(x_0,y_0)$ = $\frac{-1}{Slopeoftangent}=-\frac{a^2{y}_0}{b^2{x}_0}$

Hence, the equation of the normal to Eq. (i).at $(x_0,y_0)$ is

$y-y_0=-\frac{a^2{y}_0}{b^2{x}_0}(x-x_0)\Rightarrow \frac{y-y_0}{a^2{y}_0}=-\frac{x-x_0}{b^2{x}_0}\Rightarrow \frac{y-y_0}{a^2{y}_0}+\frac{x-x_0}{b^2{x}_0}=0$