Question

The coordinates of the point P on the curve $y^2=2x^3$,at which the tangent is perpendicular to the line 4x - 3 y + 2 = 0, are given by

• A. (2, 4)
• B. $(1,\sqrt{2})$
• C. $(\frac{1}{2},-\frac{1}{2})$
• D. $(\frac{1}{8},-\frac{1}{16})$

Answer 1

The correct option is D $(\frac{1}{8},-\frac{1}{16})$

Differentiating $y^2=2x^3$, we get
2y$\frac{dy}{dx}=6x^2\Rightarrow \frac{dy}{dx}=\frac{3x^2}{y}$
The slope of the line 4x - 3y +2 = 0 is $\frac{4}{3}$. Therefore, the coordinates of P(x, y) must satisfy
$\frac{3x^2}{y}.\frac{4}{3}$ = -1$\Rightarrow 4x^2=-y$
Also, $y^2=2x^3$ . Solving these, we get x = $\frac{1}{8}$and y = -$\frac{1}{16}$ (clearly x $\ne 0)$