Question

Let \[P\left( 4,-4 \right)\] and \[Q\left( 9,6 \right)\] be two points on a parabola \[{{y}^{2}}=4x\] and X be a point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of \[\Delta PXQ\] is maximum. Then this maximum area is:
\[\begin{align}
  & A)\dfrac{125}{4} \\
 & B)\dfrac{125}{2} \\
 & C)\dfrac{625}{4} \\
 & D)\dfrac{75}{2} \\
\end{align}\]

Answer 1

Hint: We know that a general point on the parabola \[{{y}^{2}}=4ax\] is \[X\left( a{{t}^{2}},2at \right)\]. We know that the area of triangle whose vertices are \[P\left( {{x}_{1}},{{y}_{1}} \right)\] , \[Q\left( {{x}_{2}},{{y}_{2}} \right)\] and \[R\left( {{x}_{3}},{{y}_{3}} \right)\], then the area of triangle \[\Delta PXQ\] is equal to \[\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|\]. We know that the value of \[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=a\left( ei-hf \right)-b\left( di-fg \right)+c\left( dh-ge \right)\]. By using these concepts, we can find the value of maximum area.

Complete step-by-step solution:
From the question, it was given that \[P\left( 4,-4 \right)\] and \[Q\left( 9,6 \right)\] be two points on a parabola \[{{y}^{2}}=4x\] and X be a point on the arc POQ of this parabola, where O is the vertex of this parabola.
Now let us compare \[{{y}^{2}}=4x\] with \[{{y}^{2}}=4ax\]. It is clear that the value of a is equal to 1.
We know that a general point on the parabola \[{{y}^{2}}=4ax\]is \[X\left( a{{t}^{2}},2at \right)\].
Now we should find a general point X on the parabola \[{{y}^{2}}=4x\].
So, it is clear that a general point on the parabola \[{{y}^{2}}=4ax\] is \[X\left( {{t}^{2}},2t \right)\].
Now we should find the area of triangle \[\Delta PXQ\] whose vertices are \[P\left( 4,-4 \right)\] , \[Q\left( 9,6 \right)\] and \[X\left( {{t}^{2}},2t \right)\].

We know that the area of triangle whose vertices are \[P\left( {{x}_{1}},{{y}_{1}} \right)\] , \[Q\left( {{x}_{2}},{{y}_{2}} \right)\] and \[R\left( {{x}_{3}},{{y}_{3}} \right)\], then the area of triangle \[\Delta PXQ\] is equal to \[\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|\].
By using this formula, we should find the area of triangle \[\Delta PXQ\] whose vertices are \[P\left( 4,-4 \right)\] , \[Q\left( 9,6 \right)\]and \[X\left( {{t}^{2}},2t \right)\].
Let us assume this area is equal to A.
\[\Rightarrow A=\dfrac{1}{2}\left| \begin{matrix}
   4 & -4 & 1 \\
   9 & 6 & 1 \\
   {{t}^{2}} & 2t & 1 \\
\end{matrix} \right|\]
We know that the value of \[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=a\left( ei-hf \right)-b\left( di-fg \right)+c\left( dh-ge \right)\].
\[\begin{align}
  & \Rightarrow A=\left( \dfrac{4\left( (6)(1)-(1)(2t) \right)-(-4)\left( (9)(1)-(1)({{t}^{2}}) \right)+1\left( (9)(2t)-(6)({{t}^{2}}) \right)}{2} \right) \\
 & \Rightarrow A=\left( \dfrac{4\left( 6-2t \right)+4\left( 9-{{t}^{2}} \right)+\left( 18t-6{{t}^{2}} \right)}{2} \right) \\
 & \Rightarrow A=\left( \dfrac{24-8t+36-4{{t}^{2}}+18t-6{{t}^{2}}}{2} \right) \\
 & \Rightarrow A=\left( \dfrac{-10{{t}^{2}}+10t+60}{2} \right) \\
 & \Rightarrow A=-5{{t}^{2}}+5t+30 \\
\end{align}\]
Now we should find the maximum value of A.
Now let us compare A with \[a{{x}^{2}}+bx+c\]. Then we get the value of a is equal to -5, the value of b is equal to 5 and the value of c is equal to 30.
We know that the maximum value of \[a{{x}^{2}}+bx+c\](if \[a<0\]) is equal to \[\dfrac{4ac-{{b}^{2}}}{4a}\] at \[x=\dfrac{-b}{2a}\].
Now by substituting the values if a, b and c we can find the maximum value of A.
Let us assume the maximum value of A is equal to \[{{A}_{\max }}\].
\[\begin{align}
& \Rightarrow {{A}_{\max }}=\dfrac{4\left( -5 \right)\left( 30 \right)-{{\left( 5 \right)}^{2}}}{4\left( -10 \right)} \\
 & \Rightarrow {{A}_{\max }}=\dfrac{-600-25}{-20} \\
 & \Rightarrow {{A}_{\max }}=\dfrac{625}{20} \\
 & \Rightarrow {{A}_{\max }}=\dfrac{125}{4} \\
\end{align}\]
So, it is clear that maximum area of \[\Delta PXQ\] is equal to \[\dfrac{125}{4}\].
Hence, option A is correct.

Note: Students may have a misconception that the value of \[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right|=a\left( ei-hf \right)+b\left( di-fg \right)+c\left( dh-ge \right)\]. If this misconception is followed, then we cannot get the correct value of maximum area of \[\Delta PXQ\]. So, this misconception should be avoided. So, students should have a clear view of this concept.