Question

If the tangent at any point $P(4m^2,8m^3)$ on the curve $x^3-y^2=0$ is a normal to itself at another point $Q,$ then the value(s) of $m$ is/are

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. $\sqrt{\frac{2}{9}}$
• D. $-\sqrt{\frac{2}{9}}$

Answer 1

Answer: (C), (D)

$-\sqrt{\frac{2}{9}}$

Here $y^2=x^3\cdots \left(1\right)$
$\Rightarrow 2y\frac{dy}{dx}=3x^2$
$\therefore$ Slope of tangent $={\left(\frac{3x^2}{2y}\right)}_{(4m^2,8m^3)}=3m$
Therefore, equation of tangent is $y=3mx-4m^3\cdots \left(2\right)$
For another point, solving equations $\left(1\right)$ and $\left(2\right),$ we get
$x^3={(3mx-4m^3)}^2$
$\Rightarrow x=4m^2,m^2$
$\therefore P(4m^2,8m^3)$ and $Q(m^2,-m^3)$
$\Rightarrow$ Slope of the tangent at $Q,$
${\left(\frac{dy}{dx}\right)}_{(m^2,-m^3)}={\left(\frac{3x^2}{2y}\right)}_{(m^2,-m^3)}=-\frac{3}{2}m$
$\Rightarrow$ Slope of the normal at $Q=\frac{2}{3m}$
Since tangent and normal coincide, we have
$\frac{2}{3m}=3m$
$\Rightarrow m^2=\frac{2}{9}$
$\Rightarrow m=\pm \sqrt{\frac{2}{9}}$