Question

The abcissa of the point on the curve $a{y}^2=x^3$ the normal at which cuts off equal intercepts from the axes is -

• A. $1$
• B. $\frac{4a}{3}$
• C. $3$
• D. $\frac{4a}{9}$

Answer 1

The correct option is D $\frac{4a}{9}$

Let point $(x_1,y_1)$ on the curve $a{y}^2=x^3$
$\Rightarrow 2ay\times \frac{dy}{dx}=3x^2$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x_1,y_1}=\frac{3x_1^2}{2a{y}_1}$
Thus slope of Normal is$=-{\left(\frac{dy}{dx}\right)}_{x_1,y_1}=-\frac{2a{y}_1}{3x_1^2}$
For equal intercept slope of normal should be $-1$,
$-\frac{2a{y}_1}{3x_1^2}=-1$
$\Rightarrow 3x_1^2=2a{y}_1$ ..........(1)
Also point lies on curve $a{y}_1^2=x_1^3$.......(2)
Solve (1) and (2), we get
$x_1=\frac{4a}{9}$
Hence, option 'D' is correct.