Question

Coordinates of the point P on the curve y² = 2x³ the tangent at which is perpendicular to the line 4x – 3y + 2 = 0 are given by

• A. (2, 4)
• B. (0, 0)
• C. $\left( \frac{1}{8}, - \frac{1}{16} \right)$
• D. None of these

Answer 1

Answer: (C)

$\left( \frac{1}{8}, - \frac{1}{16} \right)$

Answer 2

y² = 2x³

2y$\frac{dy}{dx}$ = 6x² Ž $\frac{dy}{dx}$ = $\frac{3x^{2}}{y}$

tangent is ^ to 4x – 3y + 2 = 0

Ž slope of this line = 4/3

Q both are ^ \ $\frac{3x^{2}}{y}$×$\frac{4}{3}$ = –1 Ž 4x² = –y

solving y² = 2x³ and 4x² = –y

we get x = 0 or x = 1/8

when x = 0 we get y = 0, but at (0, 0) slope of tangent $\frac{3x^{2}}{y}$ from (i) does not exist

\ (0, 0) does not lie on the curve

Now when x = $\frac{1}{8}$, y = –$\frac{1}{16}$. Point $\left( \frac{1}{8}, - \frac{1}{16} \right)$ lies on the curve also. So it is the required point.