For the curve y² = 2x³ - 7, the slope of the normal at (2, 3) is:

For the curve $y^2 = 2x^3 - 7$, the slope of the normal at $(2, 3)$ is:

(a)$4$
(b)$\frac{1}{4}$
(c)$-4$
(d)$-\frac{1}{4}$

Answer

Correct option: (d)

Answer: (d) $-\frac{1}{4}$ $2y\frac{dy}{dx} = 6x^2$, so $\frac{dy}{dx} = \frac{3x^2}{y}$. At $(2,3)$: slope of tangent $= \frac{12}{3} = 4$, so slope of normal $= -\frac{1}{4}$.