Question

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

Answer 1

The slope of the tanget at $(\sqrt{2}a,b)$ to the curve $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

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

$\Rightarrow y^{\prime }=\frac{b^2}{a^{2}y}{|}_{\sqrt{2}a,b}=\frac{b^2\sqrt{2}a}{a^{2}b}=\frac{b\sqrt{2}}{a}$

The equation of the tangent:

$y-b=\frac{b\sqrt{2}}{a}(x-\sqrt{2}a)$ [Using point-slope form: $y-y_1=m(x-x_1)$]

$ay-ab=b\sqrt{2}x-2ab$

or $b\sqrt{2}x-ay-ab=0$

The slope of the normal $=\frac{-1}{\frac{dy}{dx}}$

$=\frac{-1}{\frac{b\sqrt{2}}{a}}=-\frac{a}{b\sqrt{2}}$

Equation of normal:

$y-b=\frac{-a}{b\sqrt{2}}(x-\sqrt{2}a)$

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

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