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Question: If the parabolas \({{y}^{2}}=4b\left( x-c \right)\) and \({{y}^{2}}=8ax\) have a common normal, othe...

If the parabolas y2=4b(x−c){{y}^{2}}=4b\left( x-c \right) and y2=8ax{{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).
(A)(1,1,0) (B)(12,2,3) (C)(12,2,0) (D)(1,1,3) \begin{aligned} & \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{aligned}

Explanation

Solution

Hint: We solve this question by first finding the equations of normal for both curves using the formula y=mx−2am−am3y=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 y2=4b(x−c){{y}^{2}}=4b\left( x-c \right) and y2=8ax{{y}^{2}}=8ax.

Now, let us consider the equation of normal for a parabola with equation y2=4ax{{y}^{2}}=4ax,
y=mx−2am−am3y=mx-2am-a{{m}^{3}}
Using this formula, we can write equation of normal for y2=4b(x−c){{y}^{2}}=4b\left( x-c \right) as,
y=m(x−c)−2bm−bm3y=m\left( x-c \right)-2bm-b{{m}^{3}}
Using the above formula, we can write the equation of normal for y2=8ax{{y}^{2}}=8ax as,
y=mx−2(2a)m−(2a)m3 y=mx−4am−2am3 \begin{aligned} & y=mx-2\left( 2a \right)m-\left( 2a \right){{m}^{3}} \\\ & y=mx-4am-2a{{m}^{3}} \\\ \end{aligned}
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.
⇒m(x−c)−2bm−bm3=mx−4am−2am3 ⇒mx−mc−2bm−bm3=mx−4am−2am3 ⇒2am3−bm3=mc+2bm−4am ⇒(2a−b)m3=(c+2b−4a)m \begin{aligned} & \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{aligned}
As the x-axis is not included, m is not equal to zero. So,
⇒(2a−b)m2=(c+2b−4a) ⇒m2=c+2b−4a2a−b ⇒m2=c−2(2a−b)2a−b ⇒m2=c2a−b−2 \begin{aligned} & \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{aligned}
As, square of a number is always greater than zero, we get
⇒c2a−b−2>0\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)
⇒02(1)−1−2=0−2=−2<0\Rightarrow \dfrac{0}{2\left( 1 \right)-1}-2=0-2=-2<0
Does not satisfy the condition.
When (a, b, c) = (12,2,3)\left( \dfrac{1}{2},2,3 \right)
⇒32(12)−2−2=3−1−2=−3−2=−5<0\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) = (12,2,0)\left( \dfrac{1}{2},2,0 \right)
⇒02(12)−2−2=0−2=−2<0\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)
⇒32(1)−1−2=3−2=1>0\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+amy=mx+\dfrac{a}{m}, but it is wrong as that is the equation of the tangent to the parabola not for normal.