Question

If the normal to ${{y}^{2}}=12x$ at \[\left( 3,6 \right)\] meets the parabola again in $\left( 27,-18 \right)$ and the circle on the normal chord as the diameter is
(a) \[{{x}^{2}}+{{y}^{2}}+30x+12y-27=0\]
(b) \[{{x}^{2}}+{{y}^{2}}+30x+12y+27=0\]
(c) \[{{x}^{2}}+{{y}^{2}}-30x-12y-27=0\]
(d) \[{{x}^{2}}+{{y}^{2}}-30x+12y-27=0\]

Answer 1

Hint: We have to confirm the endpoints of the normal chord, for which we can use the parametric form of a point lying on the parabola \[\left( a{{t}^{2}},2at \right)\] and also the equation of the normal $y+tx=2at+a{{t}^{3}}$.

Complete step-by-step solution -
Let us consider the parabola given in the question, ${{y}^{2}}=12x$. On comparing it with the general equation of the parabola ${{y}^{2}}=4ax$ we get,
\[\begin{align}
  & 4a=12 \\
 & a=3 \\
\end{align}\]
It is given in the question that the point \[\left( 3,6 \right)\] lies on the parabola. We know that the parametric form of a point lying on the parabola is given by \[\left( a{{t}^{2}},2at \right)\]. So, on comparing the \[x\] coordinate, we get
\[\begin{align}
  & a{{t}^{2}}=3 \\
 & \therefore a=3 \\
 & 3\times {{t}^{2}}=3 \\
 & {{t}^{2}}=1 \\
\end{align}\]
Now, on comparing the $y$ coordinate, we get
\[\begin{align}
  & 2at=6 \\
 & \therefore a=3 \\
 & 2\times 3\times t=6 \\
 & 6t=6 \\
 & t=1 \\
\end{align}\]
Therefore, we get the value as \[t=1\].
Now, we know that the equation of the normal is given by $y+tx=2at+a{{t}^{3}}$. Substituting the values of \[a=3\] and \[t=1\] in it, we get
$\begin{align}
  & y+x=2\times 3\times 1+3\times {{1}^{3}} \\
 & y+x=6+3 \\
 & y+x=9 \\
 & y=9-x \\
\end{align}$
It is known that this normal intersects the parabola ${{y}^{2}}=12x$, so they must satisfy the equations of each other. Substituting \[y=9-x\] in ${{y}^{2}}=12x$, we get
${{\left( 9-x \right)}^{2}}=12x$
Since we know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\], we get
\[\begin{align}
  & 81-18x+{{x}^{2}}=12x \\
 & {{x}^{2}}-30x+81=0 \\
 & {{x}^{2}}-27x-3x+81=0 \\
 & x\left( x-27 \right)-3\left( x-27 \right)=0 \\
 & \left( x-3 \right)\left( x-27 \right)=0 \\
\end{align}\]
Therefore, we get the value as \[x=3\] or \[x=27\].
Now, from the question, it is clear that the point \[\left( 3,6 \right)\] already lies on the normal. So, using \[x=27\]to find the other point, we get
$\begin{align}
  & y=9-x \\
 & y=9-27 \\
 & y=-18 \\
\end{align}$
So, we get the point as $\left( 27,-18 \right)$. Since the circle is formed on the diameter which is the normal chords, the ends of the diameter are \[\left( 3,6 \right)\] and $\left( 27,-18 \right)$.
Thus, by using the midpoint formula given by $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$, we can get the center of the circle as,
\[\begin{align}
  & \left( \dfrac{3+27}{2},\dfrac{6-18}{2} \right) \\
 & \left( \dfrac{30}{2},\dfrac{-12}{2} \right) \\
 & \left( 15,-6 \right) \\
\end{align}\]
Now, we can plot a graph and consider the figure below.

We can find the radius by using the distance formula given by \[\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\]. Referring the figure for coordinates, we get the radius as,
\[\begin{align}
  & \sqrt{{{\left( 15-3 \right)}^{2}}+{{\left( -6-6 \right)}^{2}}} \\
 & \sqrt{{{\left( 12 \right)}^{2}}+{{\left( -12 \right)}^{2}}} \\
 & \sqrt{144+144} \\
 & \sqrt{288} \\
\end{align}\]
Thus the equation of the circle can be obtained using the general form ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$. We have the coordinates of centre \[\left( a,b \right)=\left( 15,-6 \right)\] and radius \[r=\sqrt{288}\]. So, the equation of the circle can be formulated as,
${{\left( x-15 \right)}^{2}}+{{\left( y+6 \right)}^{2}}={{\left( \sqrt{288} \right)}^{2}}$
Since we know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] and also \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], we get
$\begin{align}
  & {{x}^{2}}-30x+225+{{y}^{2}}+12y+36=288 \\
 & {{x}^{2}}+{{y}^{2}}-30x+12y+261=288 \\
 & {{x}^{2}}+{{y}^{2}}-30x+12y-27=0 \\
\end{align}$
Therefore, we get the circle as ${{x}^{2}}+{{y}^{2}}-30x+12y-27=0$ and option (d) is the correct answer.

Note: An alternative approach can be used by substituting the points \[\left( 3,6 \right)\] and $\left( 27,-18 \right)$, one by one in all the given options because if they are the ends of the diameter they must satisfy the equation of the circle and there is only a unique circle ${{x}^{2}}+{{y}^{2}}-30x+12y-27=0$ satisfying both the points.