The equation of the normal to the curve

Solution:

Given that the normal to the curve $3 x^{2}-y^{2}=8$ which is parallel to $x+3 y=k$.

Let $(a, b)$ be the point of intersection of both the curve.

$\Rightarrow 3 a^{2}-b^{2}=8 \ldots(1)$

and $a+3 b=k \ldots(2)$

Now, $3 x^{2}-y^{2}=8$

On differentiating w.r.t. $x$,

$6 x-2 y \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=\frac{3 x}{y}$

Slope of the tangent at $(a, b)=\frac{3 a}{b}$

Slope of the normal at $(a, b)=\frac{-b}{3 a}$

Slope of normal = Slope of the line

$\Rightarrow \frac{-b}{3 a}=\frac{-1}{3}$

$\Rightarrow \mathrm{b}=\mathrm{a} \ldots .(3)$

Put (3) in (1),

$3 a^{2}-a^{2}=8$

$\Rightarrow 2 a^{2}=8$

$\Rightarrow a=\pm 2$

Case: 1

When $a=2, b=2$

$\Rightarrow x+3 y=k$

$\Rightarrow k=8$

Case: 2

When $a=-2, b=-2$

$\Rightarrow x+3 y=k$

$\Rightarrow k=-8$

From both the cases,

The equation of the normal to the curve $3 x^{2}-y^{2}=8$ which is parallel to $x+3 y=k$ is $x+3 y=\pm 8$.