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Question: The equation of the normal to the parabola \[{{x}^{2}}=8y\] whose slope is \[\dfrac{1}{m}\] A.\[y=...

The equation of the normal to the parabola x2=8y{{x}^{2}}=8y whose slope is 1m\dfrac{1}{m}
A.y=mx2mm3y=mx-2m-{{m}^{3}}
B.x=mx4m2m3x=mx-4m-2{{m}^{3}}
C.x=my4m2m3x=my-4m-2{{m}^{3}}
D.4y=mx+4m2m34y=mx+4m-2{{m}^{3}}

Explanation

Solution

Hint: Differentiate the equation of the parabola x2=8y{{x}^{2}}=8y with respect to x, we will get the slope of the tangent. We know that the product of slope of two perpendicular lines is -1. Since tangent and normal are perpendicular to each other. So, Slopeoftangent !!×!! Slope of Normal = -1\text{Slope}\,\text{of}\,\text{tangent }\\!\\!\times\\!\\!\text{ Slope of Normal = -1} . Using this, we get the slope of normal and make it equal to 1m\dfrac{1}{m} . Now, get the value of x. Put the value of x and in the equation x2=8y{{x}^{2}}=8y and get the value of y. Using the formula, (yy1)=slope(xx1)\left( y-{{y}_{1}} \right)=\text{slope}\left( x-{{x}_{1}} \right) get the equation of the normal.

Complete step-by-step answer:
According to the question, it is given that,
Slope of normal = 1m\dfrac{1}{m} ………………………(1)
First of all, we have to get the slope of the tangent. For that, we have to differentiate the equation of the parabola.
Differentiating the equation of the parabola x2=8y{{x}^{2}}=8y with respect to x, we get
d(x2)dx=d(8y)dx\dfrac{d({{x}^{2}})}{dx}=\dfrac{d(8y)}{dx}………………….(2)
We know the formula, dxndx=nxn1\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}} .
Now, using this formula in equation (2), we get

& \dfrac{d({{x}^{2}})}{dx}=\dfrac{d(8y)}{dx} \\\ & \Rightarrow 2{{x}^{2-1}}=8.\dfrac{dy}{dx} \\\ & \Rightarrow 2x=8.{{m}_{T}} \\\ & \Rightarrow \dfrac{2x}{8}={{m}_{T}} \\\ \end{aligned}$$ $$\Rightarrow \dfrac{x}{4}={{m}_{T}}$$ ……………………….(3) Here, $${{m}_{T}}$$ is the slope of the tangent. We know that the product of slope of two perpendicular lines is -1. Since tangent and normal are perpendicular to each other. So, $$\text{Slope}\,\text{of}\,\text{tangent }\\!\\!\times\\!\\!\text{ Slope of Normal = -1}$$ ………………………….(4) Putting the value of $${{m}_{T}}$$ from equation (3) in equation (4), $$\text{Slope}\,\text{of}\,\text{tangent }\\!\\!\times\\!\\!\text{ Slope of Normal = -1}$$ $$\Rightarrow {{m}_{T}}\text{ }\\!\\!\times\\!\\!\text{ Slope of Normal = -1}$$ …………………….(5) $$\Rightarrow {{m}_{T}}\text{ }\\!\\!\times\\!\\!\text{ Slope of Normal = -1}$$ Putting the value of $${{m}_{T}}$$ from equation (3) in equation (5), we get $$\Rightarrow {{m}_{T}}\text{ }\\!\\!\times\\!\\!\text{ Slope of Normal = -1}$$ $$\Rightarrow \dfrac{x}{4}\text{ }\\!\\!\times\\!\\!\text{ Slope of Normal = -1}$$ $$\Rightarrow \text{Slope of Normal = }\dfrac{-4}{x}$$ ……………………….(6) From equation (1), we have the slope of normal which is equal to $$\dfrac{1}{m}$$ . Now, putting the value of slope of the normal in equation (6), $$\Rightarrow \text{Slope of Normal = }\dfrac{-4}{x}$$ $$\Rightarrow \dfrac{1}{m}\text{ = }\dfrac{-4}{x}$$ $$\Rightarrow x=-4m$$ ………………….(7) Now, putting the values of x from equation (7) in the equation of the parabola $${{x}^{2}}=8y$$ , we get $$\begin{aligned} & {{x}^{2}}=8y \\\ & \Rightarrow {{(-4m)}^{2}}=8y \\\ & \Rightarrow 16{{m}^{2}}=8y \\\ & \Rightarrow 2{{m}^{2}}=y \\\ \end{aligned}$$ Now, we have got the values of x and y. The value of x and y are $$-4m$$ and $$2{{m}^{2}}$$ respectively. Now, we have to find the equation of the normal at the point $$\left( -4m,2{{m}^{2}} \right)$$ . Equation of a is given by $$\left( y-{{y}_{1}} \right)=\text{slope}\left( x-{{x}_{1}} \right)$$ ………………………(8) Here, $$\left( {{x}_{1}},{{y}_{1}} \right)=\left( -4m,2{{m}^{2}} \right)$$ and slope of normal = $$\dfrac{1}{m}$$ . Now, putting the values of $${{x}_{1}}$$ and $${{y}_{1}}$$ in equation (8), we get $$\begin{aligned} & \left( y-2{{m}^{2}} \right)=\dfrac{1}{m}\left( x-(-4m) \right) \\\ & \Rightarrow m\left( y-2{{m}^{2}} \right)=\left( x+4m \right) \\\ & \Rightarrow my-2{{m}^{3}}=x+4m \\\ & \Rightarrow my-4m-2{{m}^{3}}=x \\\ \end{aligned}$$ Hence, the correct option of (C). Note: In this equation, one may write $$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{1}{m} \\\ & \Rightarrow \dfrac{x}{4}=\dfrac{1}{m} \\\ & \Rightarrow x=\dfrac{4}{m} \\\ \end{aligned}$$ This is wrong. According to the question, it is given that $$\dfrac{1}{m}$$ is the slope of the normal and $$\dfrac{dy}{dx}$$ is the slope of the tangent. So, $$\dfrac{1}{m}$$ cannot be equal to $$\dfrac{dy}{dx}$$ .