Question

Find the equation of normal to the parabola ${{y}^{2}}=4x$ along the minimum distance between the parabola and the circle ${{x}^{2}}+{{y}^{2}}-6x+8=0$.

Answer 1

Hint: In this question, we are given the equation of the parabola and the circle. We can use the parametrization of the parabola and the center of the circle. A normal to the parabola along the minimum distance will also be normal to the circle and hence pass through its center. Therefore, we will get an equation from which we can obtain the value of the parameter and hence the equation of the normal.

Complete step-by-step answer:
The equation of a general parabola is
${{y}^{2}}=4ax...............(1.1)$
and any point on it can be parametrized as $\left( a{{t}^{2}},2at \right)$. Comparing equation (1.1) to that given in the question which is ${{y}^{2}}=4x$, we find that a=1……………….(1.2)
Therefore, any point on the parabola can be parametrized as $\left( {{t}^{2}},2t \right)....................(1.3)$
Also, the equation of the normal to the parabola ${{y}^{2}}=4ax$ at a point with parameter t is given by
$y=-tx+a{{t}^{3}}+2at...........(1.3a)$
Using equation (1.2), the normal to the parabola ${{y}^{2}}=4x$ at a point with parameter t can be written as
$y=-tx+{{t}^{3}}+2t....................(1.4)$
Now, the equation
${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$
represents a circle with center $\left( -g,-f \right)$ …………………….. (1.5)
The equation of the circle is given to be
${{x}^{2}}+{{y}^{2}}-6x+8=0$

Comparing it with equation (1.5), we find that here $g=-3$ and $f=0$. Therefore the center of the given circle should be $\left( -\left( -3 \right),0 \right)=\left( 3,0 \right)$………….(1.6)
Now, a normal to the parabola along the minimum distance will also be normal to the circle and hence pass through its center. Therefore, (3.0) should satisfy equation (1.4), therefore,
$\begin{aligned} & 0=-t\times 3+{{t}^{3}}+2t \\\ & \Rightarrow {{t}^{3}}=t\Rightarrow t=0,\pm 1...............(1.7) \\\ \end{aligned}$
Therefore, using the value of t from equation (1.7) in equation (1.4), we get the equations of the normal as
For t=0
$\begin{aligned} & y=-0\times x+{{0}^{3}}+2\times 0 \\\ & \Rightarrow y=0....................(1.8) \\\ \end{aligned}$
For t=1,
$\begin{aligned} & y=-x+1+2 \\\ & \Rightarrow y+x=3....................(1.9) \\\ \end{aligned}$
For t=-1
$\begin{aligned} & y=-(-1)x+{{\left( -1 \right)}^{3}}-2 \\\ & \Rightarrow y=x-1-2 \\\ & \Rightarrow y=x-3....................(1.10) \\\ \end{aligned}$
Thus, the required answers to this question are $y=0$, $y+x=3$ and $y=x-3$.

Note: We should note that as equation (1.4) is a cubic equation, we should get three values of t and hence 3 normals satisfying the given condition. Also, if we parametrize the parabola by $\left( {{m}^{2}},2m \right)$, then the equation of the normal would become $y=mx-a{{m}^{3}}-2am$ in equation (1.3a). However, the obtained normals would still be the same.