Question

The point(s) on the curve $3y=6x-5x^3$ except origin at which the normal drawn also passes through origin is/are

• A. $(1,\frac{1}{3})$
• B. $(1,-\frac{1}{3})$
• C. $(\frac{-3}{5},\frac{21}{25})$
• D. $(\sqrt{\frac{3}{5}},\sqrt{\frac{3}{5}})$

Answer 1

Answer: (A), (D)

$(\sqrt{\frac{3}{5}},\sqrt{\frac{3}{5}})$

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.
$\Rightarrow 3y_1=6x_1-5x_1^3\cdots \left(ii\right)$
From equations $\left(i\right)$ and $\left(ii\right)$, we obtain
$\frac{3}{5x_1^2-2}=-5x_1^2+6$
Let $a=x_1^2$
$3=-25a^2+30a+10a-12$
$5a^2-8a+3=0$
$\Rightarrow a=1,\frac{3}{5}$
$\therefore x_1=\pm 1$ or $\pm \sqrt{\frac{3}{5}}$
$\Rightarrow x_1=1$ and $y_1=\frac{1}{3}$
$\Rightarrow x_1=-1$ and $y_1=-\frac{1}{3}$
$\Rightarrow x_1=\sqrt{\frac{3}{5}}$ and $y_1=\frac{1}{3}(\frac{6\sqrt{3}}{\sqrt{5}}-\frac{3\sqrt{3}}{\sqrt{5}})=\sqrt{\frac{3}{5}}$
$\Rightarrow x_1=-\sqrt{\frac{3}{5}}$ and $y_1=\frac{1}{3}(-\frac{6\sqrt{3}}{\sqrt{5}}+\frac{3\sqrt{3}}{\sqrt{5}})=-\sqrt{\frac{3}{5}}$
Hence, the required points are $(1,\frac{1}{3}),(-1,\frac{-1}{3}),(\sqrt{\frac{3}{5}},\sqrt{\frac{3}{5}}),(-\sqrt{\frac{3}{5}},-\sqrt{\frac{3}{5}})$