The equation of the normal to the curve $3 x^2-y^2=8$, which is parallel to the line $x+3 y=10$, is
The equation of the normal to the curve $3 x^2-y^2=8$, which is parallel to the line $x+3 y=10$, is
$x+3 y+6=0$
$x+3 y-3=0$
$x+3 y+8=0$
$x+3 y-4=0$
Solution
$3 x^2-y^2=8$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
& 6 x-2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
\therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 x}{y}
\end{aligned}$
$\therefore \quad$ Slope of the tangent to the curve is $\frac{3 x}{y}$.
$\therefore \quad$ Slope of the normal is $\frac{-y}{3 x}$.
It is parallel to line $x+3 y=10 \Rightarrow$ slope $=-\frac{1}{3}$
$\therefore \quad \frac{-y}{3 x}=\frac{-1}{3} \Rightarrow x=y$
$\therefore \quad$ When $x=y$, equation of the curve becomes
$\therefore \quad 3 x^2-x^2=8$
$\therefore \quad x^2=4$
$\therefore \quad x=2,-2 \Rightarrow y=2,-2$
$\therefore \quad(2,2)$ and $(-2,-2)$ are the points of contact of the normal and the curve.
$\therefore \quad$ Equations are $(y-2)=\frac{-1}{3}(x-2)$ or
$\begin{aligned}
& (y+2)=\frac{-1}{3}(x+2) \\
& \text { i.e., } x+3 y-8=0 \text { or } x+3 y+8=0
\end{aligned}$