Find the points on the curve 9y^2 = x^3 where the normal to the curve…

The correct answer is Option (3) → $(4,±\frac{8}{3})$

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

The given curve is $9y^2 = x^3$   ...(i)

Differentiating (i) w.r.t. x, we get $9.2y\frac{dy}{dx} = 3x^2⇒ \frac{dy}{dx}=\frac{x^2}{6y}$

∴ The slope of tangent to curve (i) at $P(x_1,y_1) =\frac{x_1^2}{6y_1}$

⇒ the slope of normal to curve (i) at $P(x_1,y_1) =-\frac{6y_1}{x_1^2}$

As the normal to the curve (i) makes equal intercepts on the coordinate axes, its slope = ±1.

$∴ -\frac{6y_1}{x_1^2}=±1⇒y_1=±\frac{x_1^2}{6}$   ...(ii)

As point $P(x_1,y_1)$ lies on the curve (i), $9y_1^2 = x_1^3$

$⇒9.(±\frac{x_1^2}{6})^2=x_1^3$   (Using (ii))

$⇒x_1^4=4x_1^3⇒x_1^3(x_1-4)=0⇒x_1=0.4$

From (ii), when $x_1 = 0, y_1 = 0$; when $x_1 = 4, y_1 = ±\frac{8}{3}$.

As the normal makes equal intercepts on the coordinate axes, it cannot pass through origin.

Therefore, the required points are $(4,\frac{8}{3})$ and $(4,-\frac{8}{3})$.