Question

If tangent to the curve $y^2=x^3$ at its point $(m^2,m^3)$ is also normal to the curve at $(M^2,M^3)$, then what is the value of mM ?

• A. - 1/9
• B. - 2/9
• C. - 1/3
• D. - 4/9

Answer 1

Answer: (D)

Equation of the given curve is $y^2=x^3$ ...(1)
On differentiating with respect to x
$2y\frac{dy}{dx}=3x^2$
$\Rightarrow \frac{dy}{dx}=\frac{3x^2}{2y}$
Now , ${\left(\frac{dy}{dx}\right)}_{\left(m^2,m^3\right)}=\frac{3m^4}{2m^3}=\frac{3}{2}m$
and ${\left(\frac{dy}{dx}\right)}_{\left(M^2,M^3\right)}=\frac{3M^4}{2M^3}=\frac{3}{2}M$
Equation of tangents at point $\left(m^2,m^3\right)$ is
$\left(y-m^3\right)=\frac{3}{2}m\left(x-m^2\right)$
$\Rightarrow 2y-2m^3=3mx-3m^3$
$\Rightarrow 3mx-2y=3m^3-2m^3$
$\Rightarrow 3mx-2y=m^3$ ....(2)
Equation of normal at point $(M^2,M^3)$ is
$\left(y-M^3\right)=-\frac{2}{3M}\left(x-M^2\right)$
$\Rightarrow 3My-3M^4=-2x+2M^2$
$\Rightarrow 2x+3My=3M^4+2M^2$ .....(3)
Since, equation (2) and (3) are same
$\Rightarrow \frac{3m}{2}=\frac{-2}{3M}=\frac{m^3}{3M^4+2M^2}$
$\Rightarrow \frac{3m}{2}=-\frac{2}{3M}$
$\Rightarrow mM=-\frac{4}{9}$