Question

The points on the curve $9y^2 = x^3$ , where the normal to the curve makes equal intercepts with the axes are

• A. $(4,±\frac{8}{3})$
• B. $(4,\frac{-8}{3})$
• C. $(4,±\frac{3}{8})$
• D. $(±4,\frac{3}{8})$

Answer 1

Answer: (A)

$(4,±\frac{8}{3})$

Answer 2

The correct answer is A:$(4±\frac{8}{3}).$
The equation of the given curve is $9y^2 = x^3 .$
Differentiating with respect to $x$, we have:
$9(2y)\frac{dy}{dx}=3x^2$
$=\frac{dy}{dx}==\frac{x^2}{6y}$
The slope of the normal to the given curve at point $(x_1,y_1)$ is
$\frac{-1}{\frac{dy}{dx}]_(x_1,y_2)}=\frac{-6y_1}{x^2_1}.$
∴ The equation of the normal to the curve at $(x_1,y_1)$ is
$y-y_1=\frac{-6y_1}{x^2_1}(x-x_1)$
$x^2_1y-x^2_1y_1=-6xy_1+6x_1y_1$
$\frac{6xy_1}{6x_1y_1+x_1^2y_1}+\frac{x_1^2y}{6x_1y_1+x_1^2y_1}$
$\frac{x}{\frac{x_1(6+x_1)}{6}}+\frac{y}{\frac{y_1(6+x_1)}{x_1}}=1$
It is given that the normal makes equal intercepts with the axes. Therefore, We have:
$\frac{x_1(6+x_1)}{6}=\frac{y_1(6+x_1)}{x_1}$
$\frac{x_1}{6}=6y_1 ....(i)$
Also, the point $(x_1,y_1)$ lies on the curve, so we have
$9y^2_1=x^3_1 ........(ii)$
From (i) and (ii), we have:
$9(\frac{x^2_1}{6})^2=x^3_1⇒\frac{x^4_1}{4}=x^3_1⇒x_1=4$
From (ii), we have:
$9y^2_1=(4)^3=64$
$⇒y^2_1=\frac{64}{9}$
$y_1=\frac{±8}{3}$
Hence, the required points are $(4±\frac{8}{3}).$
The correct answer is A.