Question

The point of the parabola \[{{y}^{2}}=64x\] which is nearest to the line \[4x+3y+35=0\] has coordinates
1) \[\left( 9,-24 \right)\]
2) \[\left( 1,81 \right)\]
3) \[\left( 4,-16 \right)\]
4) \[\left( -9,-24 \right)\]

Answer 1

Hint: In this type of question we have to use a distance formula. We know that the distance of the point \[\left( {{x}_{0}},{{y}_{0}} \right)\] from the line \[ax+by+c=0\] is given by, \[d=\dfrac{\left| a{{x}_{0}}+b{{y}_{0}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]. Also we know that if its derivative is equal to zero then the distance is minimum.

Complete step-by-step solution:
Now we have to find the coordinates of the point on the parabola \[{{y}^{2}}=64x\] which is nearest to the line \[4x+3y+35=0\]

Let us suppose that the point \[P\left( {{x}_{0}},{{y}_{0}} \right)\] is point on the parabola nearest to the line \[4x+3y+35=0\].
Now, as the point \[P\left( {{x}_{0}},{{y}_{0}} \right)\] is on the parabola \[{{y}^{2}}=64x\]
\[\Rightarrow {{y}_{0}}^{2}=64{{x}_{0}}\cdots \cdots \cdots \left( i \right)\]
As we know that the distance of the point \[\left( {{x}_{0}},{{y}_{0}} \right)\] from the line \[ax+by+c=0\] is given by, \[d=\dfrac{\left| a{{x}_{0}}+b{{y}_{0}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\].
Hence, the distance of the point \[P\left( {{x}_{0}},{{y}_{0}} \right)\] from the line \[4x+3y+35=0\] is
\[\begin{align}
  & \Rightarrow d=\dfrac{\left| 4{{x}_{0}}+3{{y}_{0}}+35 \right|}{\sqrt{{{4}^{2}}+{{3}^{2}}}} \\
 & \Rightarrow d=\dfrac{\left| 4{{x}_{0}}+3{{y}_{0}}+35 \right|}{\sqrt{25}} \\
 & \Rightarrow d=\dfrac{\left| 4{{x}_{0}}+3{{y}_{0}}+35 \right|}{5} \\
\end{align}\]
\[\Rightarrow d=\dfrac{4}{5}{{x}_{0}}+\dfrac{3}{5}{{y}_{0}}+7\]
Now by using equation \[\left( i \right)\] we will get \[{{x}_{0}}=\dfrac{{{y}_{0}}^{2}}{64}\] and hence we can write
\[\begin{align}
  & \Rightarrow d=\dfrac{4}{5}\left( \dfrac{{{y}_{0}}^{2}}{64} \right)+\dfrac{3}{5}{{y}_{0}}+7 \\
 & \Rightarrow d=\dfrac{{{y}_{0}}^{2}}{80}+\dfrac{3}{5}{{y}_{0}}+7 \\
\end{align}\]
Now as we have to find the nearest point, we have to minimise the distance and for that we differentiate the distance with respect to \[{{y}_{0}}\] and equate it to zero
\[\begin{align}
  & \Rightarrow \dfrac{d}{d{{y}_{0}}}\left( d \right)=0 \\
 & \Rightarrow \dfrac{d}{d{{y}_{0}}}\left( \dfrac{{{y}_{0}}^{2}}{80}+\dfrac{3}{5}{{y}_{0}}+7 \right)=0 \\
 & \Rightarrow \dfrac{2{{y}_{0}}}{80}+\dfrac{3}{5}=0 \\
 & \Rightarrow \dfrac{{{y}_{0}}}{40}=-\dfrac{3}{5} \\
 & \Rightarrow {{y}_{0}}=-24 \\
\end{align}\]
Now we substitute this value of \[{{y}_{0}}\] in equation \[\left( i \right)\] we get,
\[\begin{align}
  & \Rightarrow {{x}_{0}}=\dfrac{{{y}_{0}}^{2}}{64} \\
 & \Rightarrow {{x}_{0}}=\dfrac{{{\left( -24 \right)}^{2}}}{64} \\
 & \Rightarrow {{x}_{0}}=9 \\
\end{align}\]
So, the coordinates of the point is \[\left( 9,-24 \right)\]
Thus, option (1) is the correct option.

Note:In this type of question students have to remember the formula of finding the distance of a point from the line. Students have to note that they can find the minimum distance with the help of the derivatives. Also students have to take care during the calculations.