Find the equation(s) of normal(s) to the curve `3x^2-y^2=8` which is (are) parallel to the line `x+3y=4.`

Answer

Hence, the equations of the normals are `x+3 y-8=0` and `x+3 y+8=0`

Explanation

Let `P(x_{1}, y_{1})` be the foot of the required normal to the curve `3 x^{2}-y^{2}=8`.

Differentiating `3 x^{2}-y^{2}=8` w.r.t. `x`, we get
\begin{aligned}
&3 \times 2 x-2 y \frac{d y}{d x} 0 \\
&\therefore-2 y \frac{d y}{d x}=-6 \mathrm{x} \\
&\therefore \frac{d y}{d x}=\frac{3 x}{y} \\
&\therefore\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)} \\
&=\frac{3 x_{1}}{y_{1}}
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