Question

If the parabolas ${{y}^{2}}=4b\left( x-c \right)$ and ${{y}^{2}}=8ax$ have a common normal, other than x-axis, then which one of the following is a valid choice for the ordered triad (a, b, c).
$\begin{align}
  & \left( A \right)\left( 1,1,0 \right) \\
 & \left( B \right)\left( \dfrac{1}{2},2,3 \right) \\
 & \left( C \right)\left( \dfrac{1}{2},2,0 \right) \\
 & \left( D \right)\left( 1,1,3 \right) \\
\end{align}$

Answer 1

Hint: We solve this question by first finding the equations of normal for both curves using the formula $y=mx-2am-a{{m}^{3}}$. Then we equate the equations of the normal as they are common so they are the same lines. By solving them we can find a relation between slope and variables a, b and c. Then we use the inequality that square of any real number is always positive and find an inequality between a, b and c. Then we substitute the values in the options to find the triad that satisfies the inequality.

The equations of the parabolas we are given are ${{y}^{2}}=4b\left( x-c \right)$ and ${{y}^{2}}=8ax$.

Now, let us consider the equation of normal for a parabola with equation ${{y}^{2}}=4ax$,
$y=mx-2am-a{{m}^{3}}$
Using this formula, we can write equation of normal for ${{y}^{2}}=4b\left( x-c \right)$ as,
$y=m\left( x-c \right)-2bm-b{{m}^{3}}$
Using the above formula, we can write the equation of normal for ${{y}^{2}}=8ax$ as,
$\begin{align}
  & y=mx-2\left( 2a \right)m-\left( 2a \right){{m}^{3}} \\
 & y=mx-4am-2a{{m}^{3}} \\
\end{align}$
As we are given that both the parabolas have a common normal other than x-axis let us equate the normal equations of both the parabolas.
$\begin{align}
  & \Rightarrow m\left( x-c \right)-2bm-b{{m}^{3}}=mx-4am-2a{{m}^{3}} \\
 & \Rightarrow mx-mc-2bm-b{{m}^{3}}=mx-4am-2a{{m}^{3}} \\
 & \Rightarrow 2a{{m}^{3}}-b{{m}^{3}}=mc+2bm-4am \\
 & \Rightarrow \left( 2a-b \right){{m}^{3}}=\left( c+2b-4a \right)m \\
\end{align}$
As the x-axis is not included, m is not equal to zero. So,
$\begin{align}
  & \Rightarrow \left( 2a-b \right){{m}^{2}}=\left( c+2b-4a \right) \\
 & \Rightarrow {{m}^{2}}=\dfrac{c+2b-4a}{2a-b} \\
 & \Rightarrow {{m}^{2}}=\dfrac{c-2\left( 2a-b \right)}{2a-b} \\
 & \Rightarrow {{m}^{2}}=\dfrac{c}{2a-b}-2 \\
\end{align}$
As, square of a number is always greater than zero, we get
$\Rightarrow \dfrac{c}{2a-b}-2>0$
So, now let us substitute the given options and see which of them is suitable.
When (a, b, c) = (1, 1, 0)
$\Rightarrow \dfrac{0}{2\left( 1 \right)-1}-2=0-2=-2<0$
Does not satisfy the condition.
When (a, b, c) = $\left( \dfrac{1}{2},2,3 \right)$
$\Rightarrow \dfrac{3}{2\left( \dfrac{1}{2} \right)-2}-2=\dfrac{3}{-1}-2=-3-2=-5<0$
Does not satisfy the condition.
When (a, b, c) = $\left( \dfrac{1}{2},2,0 \right)$
$\Rightarrow \dfrac{0}{2\left( \dfrac{1}{2} \right)-2}-2=0-2=-2<0$
Does not satisfy the condition.
When (a, b, c) = (1, 1, 3)
$\Rightarrow \dfrac{3}{2\left( 1 \right)-1}-2=3-2=1 >0$
Satisfies the condition.
So, answer is (a, b, c) = (1, 1, 3)
So, the correct answer is Option D.

Note: A mistake that one might make while solving this problem is one might take the formula for the equation of the normal of the parabola as $y=mx+\dfrac{a}{m}$, but it is wrong as that is the equation of the tangent to the parabola not for normal.