The normal to the curve $y = x^2 - kx - 2$ at the point $(1, -4)$ is parallel to the line \(-x +

Question

The normal to the curve $y = x^2 - kx - 2$ at the point $(1, -4)$ is parallel to the line $-x + y - 1 = 0$. Find the value of $k$. Calculate the coordinates of point where this normal meets the curve again.

Accepted Answer

(i) $k=3$ (ii) $(3,-2)$

Differentiation - Tangents and Normals at a Point

Differentiation - Tangents and Normals at a Point

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