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Question

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


A
(4,±83)
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B
(4,83)
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C
(4,±38)
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D
(±4,38)
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Solution

The correct option is A $$\left(4,\pm\dfrac{8}{3}\right)$$
Let $$\left(h,k\right)$$ be the point on the curve $$9{y}^{2}={x}^{3}$$ where the normal makes equal intercept with the axes.

We know that the slope of tangent on the curve is $$\dfrac{dy}{dx}$$
$$9{y}^{2}={x}^{3}$$

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

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

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

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

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

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

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

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

Since normal is at point $$\left(h,k\right)$$

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

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

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

$$\Rightarrow\,\dfrac{x}{a}+\dfrac{y}{a}=1$$ 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 $$\left(h,k\right)=-1$$        ........$$(2)$$

From $$(1)$$ and $$(2)$$ we have

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

$$\Rightarrow\,6k={h}^{2}$$     .......$$(3)$$

Also,point $$\left(h,k\right)$$ is on the curve $$9{y}^{2}={x}^{3}$$

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

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

$$\Rightarrow\,9{k}^{2}={h}^{3}$$    .......$$(4)$$

Now, our equations are 

$$6k={h}^{2}$$     .......$$(3)$$ 

$$9{k}^{2}={h}^{3}$$    .......$$(4)$$

From $$(3)$$

$$6k={h}^{2}$$ 

$$\Rightarrow\,k=\dfrac{{h}^{2}}{6}$$

$$\Rightarrow\,k=\pm\dfrac{8}{9}$$

Put value of $$k$$ in $$(4)$$ we get

$$9{k}^{2}={h}^{3}$$

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

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

$$\Rightarrow\,\dfrac{h}{4}=1$$

$$\Rightarrow\,h=4$$

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

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

$$\Rightarrow\,{k}^{2}=\dfrac{64}{9}$$

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

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


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