Question

If the tangent at any point $P\left(4m^2,8m^3\right)$ of $x^3−y^2=0$ is also a normal to the curve $x^3−y^2=0$, then the value of $m$ is

• A. $m=\frac{\sqrt{2}}{3}$
• B. $m=−\frac{\sqrt{2}}{3}$
• C. $m=\frac{3}{\sqrt{2}}$
• D. $m=−\frac{3}{\sqrt{2}}$

Answer 1

Answer: (A), (B)

$x^3−y^2=0...(i)$
$⇒2y×\frac{dy}{dx}=3x^2$
Slope of the tangent at $P={\frac{dy}{dx}â}_p={\frac{3x^2}{2y}â}_{\left(4m^2.8m^3\right)}=3m$
$∴$ Equation of the tangent at $P$ is
$y−8m^3=3m\left(x−4m^2\right)$
or $y=3mx−4m^3...(ii)$
It cuts the curve again at point $Q$.
Solving (i) and (ii), we get $x=4m^2,m^2$
Put $x=m^2$ in equation (ii)
$⇒y=3m\left(m^2\right)−4m^3=−m^3$
$∴Q$ is $\left(m^2,−m^3\right)$
Slope of the tangent at $Q={\frac{dy}{dx}â}_{\left(m^2,−m^3\right)}=\frac{3\left(m^4\right)}{2×\left(−m^3\right)}=\frac{−3}{2}m$
Slope of the normal at $Q=\frac{1}{(−3/2)m}=\frac{2}{3m}$
Since tangent at $P$ is normal at $Q$
$⇒\frac{2}{3m}=3m$
$⇒9m^2=2$