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

The Correct Option is A

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.