Question

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

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

Answer 1

The correct option is A $(4,\pm \frac{8}{3})$

Let $(h,k)$ be the point on the curve $9y^2=x^3$ where the normal makes equal intercept with the axes.

We know that the slope of tangent on the curve is $\frac{dy}{dx}$

$9y^2=x^3$

Differentiating w.r.t.$x$ we get

$\frac{d\left(9y^2\right)}{dx}=\frac{d\left(x^3\right)}{dx}$

$\Rightarrow 9\frac{d\left(y^2\right)}{dx}=3x^2$

$\Rightarrow 18y\frac{dy}{dx}=3x^2$

$\Rightarrow \frac{dy}{dx}=\frac{x^2}{6y}$

We know that slope of tangent$\times$Slope of normal$=-1$

$\Rightarrow \frac{x^2}{6y}\times$Slope of normal$=-1$

$\Rightarrow$Slope of normal$=\frac{-6y}{x^2}$

Since normal is at point $(h,k)$

Hence slope of normal at $(h,k)$ is $\frac{-6k}{h^2}$ .....$\left(1\right)$

Equation of normal to the curve makes equal intercept with axes is

$\frac{x}{a}+\frac{y}{b}=1$

$\Rightarrow \frac{x}{a}+\frac{y}{a}=1$ where $a=b$

$\Rightarrow x+y=a$

$\Rightarrow y=-x+a$ which is of the form $y=mx+c$

where $m$ is the slope of the line

$\Rightarrow$Slope of normal$=-1$

$\Rightarrow$Slope of normal at $(h,k)=-1$ ........$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we have

$\frac{-6k}{h^2}=1$

$\Rightarrow 6k=h^2$ .......$\left(3\right)$

Also,point $(h,k)$ is on the curve $9y^2=x^3$

$\Rightarrow (h,k)$ will satisfy the equation of curve

Putting $x=h$ and $y=k$ in equation

$\Rightarrow 9k^2=h^3$ .......$\left(4\right)$

Now, our equations are

$6k=h^2$ .......$\left(3\right)$

$9k^2=h^3$ .......$\left(4\right)$

From $\left(3\right)$

$6k=h^2$

$\Rightarrow k=\frac{h^2}{6}$

$\Rightarrow k=\pm \frac{8}{9}$

Put value of $k$ in $\left(4\right)$ we get

$9k^2=h^3$

$\Rightarrow 9{\left(\frac{h^2}{6}\right)}^2=h^3$

$\Rightarrow \frac{9h^4}{36}=h^3$

$\Rightarrow \frac{h}{4}=1$

$\Rightarrow h=4$

Put the value of $h=4$ in $\left(4\right)$ we get

$9k^2=h^3={\left(4\right)}^3$

$\Rightarrow k^2=\frac{64}{9}$

$\Rightarrow k=\pm \sqrt{\frac{64}{9}}=\pm \frac{8}{3}$

Hence the required point is $(h,k)=(4,\pm \frac{8}{3})$