Question

The point $P$ of curve $y^2=2x^3$ such that the tangent at $P$ is perpendicular to the line $4x-3y+2=0$ is given by :

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

Answer 1

Answer: (D)

Given equation of curve : $y^2=2x^3...\left(i\right)$
and line : $4x-3y+2=0$
$y^2=2x^3\Rightarrow 2y\frac{dy}{dx}=6x^2$
Slope of tangent
$\frac{dy}{dx}=\frac{6x^2}{2y}=\frac{3x^2}{y}=m_1\left(say\right)$
and $4x-3y+2=0\Rightarrow -3y=-4x-2$
$3y=4x+2\Rightarrow y=\frac{4}{3}x+\frac{2}{3}$
Slope of line $=\frac{4}{3}=m_2\left(say\right)$
Now, $m_1\times m_2=-1$
$\therefore \frac{3x^2}{y}\times \frac{4}{3}=-1$
$y=-4x^2$
Squaring on both sides, we get
$y^2=16x^4...\left(ii\right)$
$equ.\left(i\right)-equ\left(ii\right):$
$2x^3=16x^4$
$\Rightarrow x=\frac{1}{8}$ or $0$
If $x=0$ then $y=0$
If $x=\frac{1}{8}$ then
$y=-4\times \frac{1}{8}\times \frac{1}{8}=-\frac{1}{16}$
Therefore, the points are $\left(0,0\right)$ and $\left(\frac{1}{8},-\frac{1}{16}\right).$