If the line $x - y + 2 = 0$ is a normal to the parabola $y^{2} - 6 y - 4 x + k = 0$ , then find $k$ .

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Solution

We have,
$x - y + 2 = 0 [s l o p e = 1] y^{2} - 6 y - 4 x + k = 0 2 y \frac{d y}{d x} - \frac{6 d y}{d x} - 4 = 0 \frac{d y}{d x} = \frac{4}{2 y - 6} - \frac{d x}{d y} = \frac{6 - 2 y}{4} = 1 y = 1 t h a t g i v e s x = - 1 f r o m w h i c h k = 1$

Hence, this is the answer.

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