Question

Find the equation of the normal to the curve \[y={{x}^{3}}+2x+6\] which are parallel to \[x+14y+4=0\].

Answer 1

Hint: We know that the slope of a tangent at \[A({{x}_{1}},{{y}_{1}})\] to a curve \[f(x)=0\] is \[\dfrac{dy}{dx}=0\]. So, we have to find the slope of \[y={{x}^{3}}+2x+6\] at \[A({{x}_{1}},{{y}_{1}})\]. This gives us the slope of tangent. Now we have to find the slope of normal. We know that the normal is a line perpendicular to slope.
We also know that a line of slope \[{{m}_{1}}\] is said to be perpendicular to a line of slope of \[{{m}_{2}}\],if \[{{m}_{1}}{{m}_{2}}=-1\]. Now we have to find the slope of \[x+14y+4=0\]. We know that the normal is parallel to \[x+14y+4=0\]. We know that if two lines are parallel, then their slopes are equal. From this we will find the coordinates of the points where the normals are drawn. Now find the equation of required normals.

Complete step-by-step answer:
We know that the slope of a tangent at \[A({{x}_{1}},{{y}_{1}})\] to a curve \[f(x)=0\] is \[\dfrac{dy}{dx}=0\].
Now we have to find the slope of a tangent at \[A({{x}_{1}},{{y}_{1}})\] to a curve \[y={{x}^{3}}+2x+6\].
Now we have to differentiate \[y={{x}^{3}}+2x+6\] with respect to x on both sides.
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d}{dx}({{x}^{3}}+2x+6) \\
 & \Rightarrow \dfrac{dy}{dx}=3{{x}^{2}}+2 \\
 & \Rightarrow {{\left( \dfrac{dy}{dx} \right)}_{\left( {{x}_{1}},{{y}_{1}} \right)}}=3x_{1}^{2}+2.....(1) \\
\end{align}\]
We know that if the slope of \[ax+by+c=0\] is m, then \[m=\dfrac{-a}{b}\].
We know that the normal is a line perpendicular to slope.
We also know that a line of slope \[{{m}_{1}}\] is said to be perpendicular to a line of slope of \[{{m}_{2}}\],if \[{{m}_{1}}{{m}_{2}}=-1\].
From equation (1), we are having the slope of tangent.
Let us assume the slope of normal is equal to \[m\grave{\ }\].
Then we get
\[mm’=-1....(2)\]
Now let us substitute equation (1) in equation (3), then we get
\[\begin{align}
  & \left( 3x_{1}^{2}+2 \right)m’ =-1 \\
 & \Rightarrow m’ =\dfrac{-1}{\left( 3x_{1}^{2}+2 \right)}....(3) \\
\end{align}\]
From the question, we were given that the normal is parallel to \[x+14y+4=0\].
Now we have to find the slope of \[x+14y+4=0\].
We know that if the slope of \[ax+by+c=0\] is m, then \[m=\dfrac{-a}{b}\].
Let us assume the slope of \[x+14y+4=0\] is equal to \[m’’ \].
\[m’’ =-\dfrac{1}{14}.....(4)\]
We know that if two lines are equal, then their slopes are equal.
So, the slope of normal is equal to \[\dfrac{-1}{14}\].
So, from equation (3) and equation (4) we get
\[\begin{align}
  & \dfrac{-1}{\left( 3x_{1}^{2}+2 \right)}=\dfrac{-1}{14} \\
 & \Rightarrow \left( 3x_{1}^{2}+2 \right)=14 \\
 & \Rightarrow 3x_{1}^{2}=12 \\
 & \Rightarrow x_{1}^{2}=4 \\
 & \Rightarrow {{x}_{1}}=\pm 2.....(5) \\
\end{align}\]
If \[A({{x}_{1}},{{y}_{1}})\] is a point on the curve \[y={{x}^{3}}+2x+6\] then
\[{{y}_{1}}=x_{1}^{3}+2{{x}_{1}}+6.......(6)\]
Case 1: (If \[{{x}_{1}}=2\] )
Now substitute \[{{x}_{1}}=2\] in equation (6), then we get
\[\begin{align}
  & \Rightarrow {{y}_{1}}={{(2)}^{3}}+2(2)+6 \\
 & \Rightarrow {{y}_{1}}=18......(7) \\
\end{align}\]
Case 2: (If \[{{x}_{1}}=-2\] )
Now substitute \[{{x}_{1}}=-2\] in equation (6), then we get
\[\begin{align}
  & \Rightarrow {{y}_{1}}={{(-2)}^{3}}+2(-2)+6 \\
 & \Rightarrow {{y}_{1}}=-6......(8) \\
\end{align}\]
So, it is clear that a normal is parallel to \[x+14y+4=0\] if the normal is drawn at (2, 18) and (-2, -6).
We know that an equation of line of slope m and passes through \[A({{x}_{1}},{{y}_{1}})\] is \[y-{{y}_{1}}=m(x-{{x}_{1}})\].
Now we have to find the equation of the line whose slope is equal to \[\dfrac{-1}{14}\] and passes through (2, 18).
\[\begin{align}
  & \Rightarrow y-18=\left( \dfrac{-1}{14} \right)(x-2) \\
 & \Rightarrow 14y-(252)=-x+2 \\
 & \Rightarrow x+14y-254=0 \\
\end{align}\]
Now we have to find the equation of the line whose slope is equal to \[\dfrac{-1}{14}\] and passes through (-2, -6).
\[\begin{align}
  & \Rightarrow y-(-6)=\left( \dfrac{-1}{14} \right)(x-(-2)) \\
 & \Rightarrow y+6=\left( \dfrac{-1}{14} \right)(x+2) \\
 & \Rightarrow 14y+84=-x-2 \\
 & \Rightarrow x+14y+86=0 \\
\end{align}\]

So, the equation of the normal to the curve \[y={{x}^{3}}+2x+6\] which are parallel to \[x+14y+4=0\] are \[x+14y-254=0\] and \[x+14y+86=0\].

Note: Students have a misconception that the slope of \[ax+by+c=0\], then \[m=\dfrac{-a}{b}\]. If this misconception is followed, then it will give wrong results. Students may also have misconception that that a line of slope \[{{m}_{1}}\] is said to be perpendicular to a line of slope of \[{{m}_{2}}\] ,if \[{{m}_{1}}{{m}_{2}}=1\]. If this misconception is followed, then it will give wrong results. These misconceptions should be avoided to have a correct solution.