Question

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

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

Answer 1

The correct option is A

$\left(1,\frac{1}{3}\right)$

The explanation for the correct option.

Step-1 : Slope of the normal of the curve:

Let $P(a,b)$ be the required point.

Given the equation of curve is $3y=6x-5x^3$

Differentiate with respect to $x$.

$$\begin{gathered} \therefore 3\frac{dy}{dx}=6-15x^2 \\ \Rightarrow \frac{dy}{dx}=2-5x^2 \end{gathered}$$

The product of the slope of two perpendicular lines is equal to $-1$.

Thus, the Slope of the tangent at $P(a,b)$$=2-5a^2$

Slope of normal $=\frac{1}{5a^2-2}$

Step-2: Equation of the normal:

Equation of normal at $P$ is computed as,

$y-b=\frac{1}{5a^2-2}(x-a)$

Since it passes through the origin. So substitute $(x,y)=(0,0)$.

$\therefore b=\frac{a}{5a^2-2}$

When we substitute all the given options in the above equation. Then, the only option $A$satisfies the above equation.

Thus, the required point is $\left(1,\frac{1}{3}\right)$.

Hence, option $A$ is correct.

The explanation for the incorrect option.

Since the option $B,C,\mathrm{and}D$ does not satisfy the equation $b=\frac{a}{5a^2-2}$.

Hence, the option $B,C,\mathrm{and}D$ are incorrect.

Hence, option $A$ is correct.