Question

The point on the curve $3y=6x-5x^3,$ at which the normal passes through origin, is

• A. $(1,\frac{1}{3})$
• B. $(\frac{1}{3},1)$
• C. $(2,-\frac{28}{3})$
• D. None of these.

Answer 1

The correct option is A $(1,\frac{1}{3})$

Let the required point be $(x_1,y_1)$

Now, $3y=6x-5x^3$

$\Rightarrow 3\frac{dy}{dx}=6-15x^2$

$\Rightarrow \frac{dy}{dx}=2-5x^2$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=2-5x_1^2$

The equation of the normal at $(x_1,y_1)$ is $y-y_1=\frac{-1}{2-5x_1^2}(x-x_1)$

If it passes through the origin, then

$0-y_1=\frac{-1}{2-5x_1^2}(0-x_1)$

$\Rightarrow y_1=\frac{-x_1}{2-5x_1^2}\cdots \left(i\right)$

Since $(x_1,y_1)$ lies on the given curve.

Therefore, $3y_1=6x_1-5x_1^3\cdots \left(ii\right)$

Solving equations $\left(i\right)$ and $\left(ii\right)$, we obtain

$\frac{3}{5x_1^2-2}=-5x_1^2+6$
Let $a=x_1^2$
$3=-25a^2+30a+10a-12$
$5a^2-8a+3=0$
$\therefore a=1$
$\Rightarrow x_1=1$ and $y_1=\frac{1}{3}$
$\Rightarrow x_1=-1$ and $y_1=-\frac{1}{3}$
Hence, the required point is $(1,\frac{1}{3})$.