Question

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

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

Answer 1

Answer: (A)

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

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

Given equation of curve is

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

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

Slope of tangent at P is $2-5{x_1}^2$

Slope of normal at P is $\frac{1}{5{x_1}^2-2}$

Equation of normal at P is

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

Since it passes through origin,

$\Rightarrow y_1=\frac{x_1}{5{x_1}^2-2}$

Here , we can solve by check options.

So ,option A satisfies above equation.

Hence $P(1,\frac{1}{3})$ is the required point.