Question

Assertion (A): The normal to the curve $a{y}^2=x^3(a\ne 0,x\ne 0)$ at a point $(x,y)$ on it makes equal intercepts on the axes, then $x=\frac{4a}{9}$.
Reason (R): The normal at $(x_1,y_1)$ on the curve $y=f\left(x\right)$ makes equal intercepts on the coordinate axes, then ${\frac{dy}{dx}|}_{(x_1,y_1)}=1$

• A. Both (A) and (R) are true and (R) is the correct explanation for (A).
• B. Both (A) and (R) are true but (R) is not the correct explanation for (A).
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true.

Answer 1

The correct option is A Both (A) and (R) are true and (R) is the correct explanation for (A).

$a{y}^2=x^3$
$2ay\frac{dy}{dx}=3x^2\Rightarrow \frac{dy}{dx}=\frac{3x^2}{2ay}$
Now, slope of normal
$=-\frac{dx}{dy}=-\frac{2ay}{3x^2}=-\frac{2a\times x^{\frac{3}{2}}}{3x^2\sqrt{a}}=-\frac{2\sqrt{a}}{3\sqrt{x}}$
For the normal to make equal intercepts on the axes, the slope has to be $\pm 1$.
Hence, $\frac{-2\sqrt{a}}{3\sqrt{x}}=\pm 1$
On squaring both sides,
$\frac{4a}{9x}=1\Rightarrow x=\frac{4a}{9}$.