Question

The abscissa of the point on the curve 3y = 6x - 5x³, the normal at which passes through the origin is
(a) 1 (b) $\frac{1}{3}$ (c) 2 (d) $\frac{1}{2}$

Answer 1

Let (h, k) be the point on the curve 3y = 6x − 5x³, the normal at which passes through the origin.

∴ 3k = 6h − 5h³ .....(1)

3y = 6x − 5x³

Differentiating both sides with respect to x, we get

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

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

∴ Slope of tangent to the given curve at (h, k) = ${\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=2-5h^2$

Slope of normal to the given curve at (h, k) = $-\frac{1}{{\left({\displaystyle \frac{dy}{dx}}\right)}_{\left(h,k\right)}}=-\frac{1}{\left(2-5h^2\right)}$

Equation of the normal to the given curve at (h, k) is


$y-k=-\frac{1}{\left(2-5h^2\right)}\left(x-h\right)$

This equation passes through the origin.

$\therefore 0-k=-\frac{1}{\left(2-5h^2\right)}\left(0-h\right)$

$\Rightarrow k=-\frac{h}{2-5h^2}$ .....(2)

From (1) and (2), we have

$-\frac{h}{2-5h^2}=\frac{6h-5h^3}{3}$

$\Rightarrow \frac{1}{5h^2-2}=\frac{6-5h^2}{3}$

$\Rightarrow 30h^2-25h^4-12+10h^2=3$

$\Rightarrow 25h^4-40h^2+15=0$

$\Rightarrow 5h^4-8h^2+3=0$

$\Rightarrow 5h^4-5h^2-3h^2+3=0$

$\Rightarrow 5h^2\left(h^2-1\right)-3\left(h^2-1\right)=0$

$\Rightarrow \left(h^2-1\right)\left(5h^2-3\right)=0$

⇒ h² − 1 = 0 or 5h² − 3 = 0

⇒ h² = 1 or $h^2=\frac{3}{5}$

⇒ h = ±1 or $h=\pm \sqrt{\frac{3}{5}}$

Thus, the abscissa of the points on the given curve, the normal at which passes through the origin are $-\sqrt{\frac{3}{5}}$, −1, 1 and $\sqrt{\frac{3}{5}}$.

Hence, the correct answer is option (a).