Find the equation of the normal lines to the curve $3 x^{2} - y^{2} = 8$ which are parallel to the line $x + 3 y = 4.$

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Solution

Given equation of the curve is:
$3 x^{2} - y^{2} = 8$
On differentiating both sides w.r.t. x, we get
$6 x - 2 y \frac{d y}{d x} = 0$
$\Rightarrow \frac{d y}{d x} = \frac{6 x}{2 y} = \frac{3 x}{y} \Rightarrow m_{1} = \frac{3 x}{y}$
and slope of normal $(m_{2}) = \frac{- 1}{m_{1}} = \frac{- y}{3 x}$
Slope of normal to the curve should be equal to the slope of line x + 3 y = 4 which is parallel to curve.
For line, $y = \frac{4 - x}{3} = \frac{- x}{3} + \frac{4}{3} \Rightarrow S l o p e o f t h e l i n e (m_{3}) = \frac{- 1}{3} ∴ m_{2} = m_{3} \Rightarrow \frac{- y}{3 x} == \frac{1}{3} \Rightarrow y = x$
On substituting the value of y in Eq. (i) we get
$3 x^{2} - x^{2} = 8 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2 a n d f o r x = - 2, y = - 2$
Thus the points at which normal to the curve are parallel to the line x + 3y = 4 are (2,2) and (-2, -2).
Required equations of normal are
$y - 2 = m_{2} (x - 2) a n d y + 2 = m_{2} (x + 2) \Rightarrow y - 2 = \frac{- 2}{6} (x - 2) a n d y + 2 = \frac{- 2}{6} (x + 2) \Rightarrow 3 y - 6 = - x + 2 a n d 3 y + 6 = - x - 2 \Rightarrow 3 y + x = + 8 a n d 3 y + x = - 8$
So, the required equations are $3 y + x = \pm 8$

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