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)
B
(4,83)
C
(4,±38)
D
(±4,38)

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 curvePutting $$x=h$$ 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 $$\left(h,k\right)=\left(4,\pm\dfrac{8}{3}\right)$$Maths

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