Question

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

Answer 1

Hint: Differentiate the equation of the parabola \[{{x}^{2}}=8y\] 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, \[\text{Slope}\,\text{of}\,\text{tangent }\!\!\times\!\!\text{ Slope of Normal = -1}\] . Using this, we get the slope of normal and make it equal to \[\dfrac{1}{m}\] . Now, get the value of x. Put the value of x and in the equation \[{{x}^{2}}=8y\] and get the value of y. Using the formula, \[\left( y-{{y}_{1}} \right)=\text{slope}\left( x-{{x}_{1}} \right)\] get the equation of the normal.

Complete step-by-step answer:
According to the question, it is given that,
Slope of normal = \[\dfrac{1}{m}\] ………………………(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 \[{{x}^{2}}=8y\] with respect to x, we get
\[\dfrac{d({{x}^{2}})}{dx}=\dfrac{d(8y)}{dx}\]………………….(2)
We know the formula, \[\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}\] .
Now, using this formula in equation (2), we get
\[\begin{align}
  & \dfrac{d({{x}^{2}})}{dx}=\dfrac{d(8y)}{dx} \\
 & \Rightarrow 2{{x}^{2-1}}=8.\dfrac{dy}{dx} \\
 & \Rightarrow 2x=8.{{m}_{T}} \\
 & \Rightarrow \dfrac{2x}{8}={{m}_{T}} \\
\end{align}\]
\[\Rightarrow \dfrac{x}{4}={{m}_{T}}\] ……………………….(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
 \[\begin{align}
  & {{x}^{2}}=8y \\
 & \Rightarrow {{(-4m)}^{2}}=8y \\
 & \Rightarrow 16{{m}^{2}}=8y \\
 & \Rightarrow 2{{m}^{2}}=y \\
\end{align}\]
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
\[\begin{align}
  & \left( y-2{{m}^{2}} \right)=\dfrac{1}{m}\left( x-(-4m) \right) \\
 & \Rightarrow m\left( y-2{{m}^{2}} \right)=\left( x+4m \right) \\
 & \Rightarrow my-2{{m}^{3}}=x+4m \\
 & \Rightarrow my-4m-2{{m}^{3}}=x \\
\end{align}\]
Hence, the correct option of (C).

Note: In this equation, one may write
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{1}{m} \\
 & \Rightarrow \dfrac{x}{4}=\dfrac{1}{m} \\
 & \Rightarrow x=\dfrac{4}{m} \\
\end{align}\]
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 \[\dfrac{dy}{dx}\] .