Question

The coordinates of the point on the curve $x^3=y(x-a{)}^2,a>0$ where the ordinate is minimum

• A. $(2a,8a)$
• B. $(-2a,\frac{-8a}{9})$
• C. $(3a,\frac{27a}{4})$
• D. $(-3a,\frac{-27a}{16})$

Answer 1

The correct option is C $(3a,\frac{27a}{4})$

The ordinates of any point on the curve is given by $y=\frac{x^3}{(x-a{)}^2}$
$\frac{dy}{dx}=\frac{x^2(x-3a)}{(x-a{)}^3}$
Now, $\frac{dy}{dx}=0\Rightarrow x=0$ or x=3a
$\frac{d^{2}y}{d{x}^2}{|{}_{x=0}=0and\frac{d^{2}y}{d{x}^2}|}_{x=3a}=\frac{72a^5}{(2a{)}^6}>0$
Hence y is minimum at x = 3a and is equal to $\frac{27a}{4}$