Question

Find the equations of the tangent and normal to the curve $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the point $(\sqrt{2}a,b).$

Answer 1

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Differentiating. w.r.t. x, we get

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

At $(\sqrt{2}a,b),\frac{dy}{dx}=\frac{b^2}{a^2}.\frac{\sqrt{2}a}{b}=\frac{\sqrt{2}b}{a}$

$\therefore$ Equation of tangent at $(\sqrt{2}a,b)is$
$y-b=\frac{\sqrt{2}b}{a}(x-\sqrt{2}a)$

$\Rightarrow ay-ab=\sqrt{2}bx-2ab\Rightarrow \sqrt{2}bx-ay-ab=0$
$Slopeofnormal=-\frac{1}{Slopeoftangent}=-\frac{1}{\frac{\sqrt{2}b}{a}}=-\frac{a}{\sqrt{2}b}$
Equation of normal at $(\sqrt{2}a,b)$ is
$y-b=-\frac{a}{\sqrt{2}b}(x-\sqrt{2}a)$

$\Rightarrow \sqrt{2}by-\sqrt{2}{b}^2=-ax+\sqrt{2}{a}^2$

$\Rightarrow ax+\sqrt{2}by-\sqrt{2}(a^2+b^2)=0$