The equation of normal to the curve 3x2 – y2 = 8 which is parallel to the line x + 3y = 8 is ______.

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The equation of normal to the curve 3x2 – y2 = 8 which is parallel to the line x + 3y = 8 is x + 3y $\pm$ 8 = 0. Explanation: Given equation of the curve is 3x2 – y2 = 8 ......(i) Differentiating both sides w.r.t. x, we get 6x - 2y * "dy"/"dx" = 0 $\Rightarrow$ "dy"/"dx" = (3x)/y (3x)/y is the slope of the tangent $\therefore$ Slope of the normal = (-1)/("dy"/"dx") = (-y)/(3x) Now x + 3y = 8 is parallel to the normal Differentiating both sides w.r.t. x, we have 1 + 3 "dy"/"dx" = 0 $\Rightarrow$ "dy"/"dx" = - $\frac{1}{3}$ $\therefore$ (-y)/(3x) = - $\frac{1}{3}$ $\Rightarrow$ y = x Putting y = x in equation (i) we get 3x2 – x2 = 8 $\Rightarrow$ 2x2 = 8 $\Rightarrow$ x2 = 4 $\therefore$ x = $\pm$ 2 and y = $\pm$ 2 So the points are (2, 2) and (– 2, – 2). Equation of normal to the given curve at (2, 2) is y – 2 = - $\frac{1}{3}$(x - 2) $\Rightarrow$ 3y – 6 = – x + 2 $\Rightarrow$ x + 3y – 8 = 0 Equation of normal at (– 2, – 2) is y + 2 = - $\frac{1}{3}$ (x + 2) $\Rightarrow$ 3y + 6 = – x – 2 $\Rightarrow$ x + 3y + 8 = 0 $\therefore$ The equations of the normals to the curve are x + 3y $\pm$ 8 = 0.

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