Question

Let A (4, – 4) and B (9, 6) be the point on the parabola, \[{{y}^{2}}=4x.\] Let C be chosen the arc AOB of the parabola, where O is the origin, such that the area of triangle ACB is maximum. Then, the area (in square units) of triangle ACB is:
\[\left( a \right)31\dfrac{3}{4}\]
\[\left( b \right)32\]
\[\left( c \right)30\dfrac{1}{2}\]
\[\left( d \right)31\dfrac{1}{4}\]

Answer 1

Hint: We are given a parabola \[{{y}^{2}}=4x.\] We can see that it is a parabola symmetric to the x-axis. We will assume that point C is at a location (x, y). Now, we know the formula of the area of the triangle, \[\text{Area}=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{7}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\] to find the area of triangle ACB. For the maximum area, \[\dfrac{d\left( \text{Area} \right)}{dx}\] must be zero. We will compute the derivative of the area to compare it to zero which will give us the point of maxima. Then putting \[x=\dfrac{1}{4}\] we will get the maximum area.

Complete step-by-step solution:
We are given a parabola defined as \[{{y}^{2}}=4x.\] We know that the parabola \[{{y}^{2}}=4ax\] is the parabola which is symmetric to the x-axis. So, our parabola \[{{y}^{2}}=4x\] is symmetric to the x-axis. Also, we have A (4, – 4) and B (9, 6) lie on the parabola.
We are asked to find C that lies on the curve AOB of the parabola such that area of triangle ACB is maximum. Let us assume the point C has coordinates as (x, y). So, we have,

As C (x, y) lies on the parabola \[{{y}^{2}}=4x\] so it must satisfy the equation of the parabola.
\[\Rightarrow {{y}^{2}}=4x\]
\[\Rightarrow y=2\sqrt{x}\]
So our point C (x, y) can be written as \[C\left( x,2\sqrt{x} \right).\] Now, for triangle ACB, we have the coordinate as \[A\left( 4,-4 \right),B\left( 9,6 \right),C\left( x,2\sqrt{x} \right).\] We know that the area of the triangle is given as,
\[\text{Area}=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{7}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\]
As we have,
\[\left( {{x}_{1}},{{y}_{1}} \right)=\left( 4,-4 \right)\]
\[\left( {{x}_{2}},{{y}_{2}} \right)=\left( 9,6 \right)\]
\[\left( {{x}_{3}},{{y}_{3}} \right)=\left( x,2\sqrt{x} \right)\]
So, we get,
\[\Rightarrow \text{Area}=\dfrac{1}{2}\left[ 4\left( 6-2\sqrt{x} \right)+9\left( 2\sqrt{x}-\left( -4 \right) \right)+x\left( -4-6 \right) \right]\]
\[\Rightarrow \text{Area}=\dfrac{1}{2}\left[ 24-8\sqrt{x}+18\sqrt{x}+36-10x \right]\]
Simplifying further, we get,
\[\Rightarrow \text{Area}=\dfrac{1}{2}\left[ 60+10\sqrt{x}-10x \right]\]
Now, as we are looking for a maximum area, so we will find the critical point. To do so, we will differentiate the area with respect to x. So,
\[\dfrac{d\left( \text{Area} \right)}{dx}=\dfrac{d\left( \dfrac{60+10\sqrt{x}-10x}{2} \right)}{dx}\]
Simplifying further, we get,
\[\Rightarrow \dfrac{d\left( \text{Area} \right)}{dx}=\dfrac{d\left( 30+5\sqrt{x}-5x \right)}{dx}\]
\[\Rightarrow \dfrac{d\left( \text{Area} \right)}{dx}=\dfrac{d\left( 30 \right)}{dx}+\dfrac{d\left( 5\sqrt{x} \right)}{dx}-\dfrac{d\left( 5x \right)}{dx}\]
After deriving, we get,
\[\Rightarrow \dfrac{d\left( \text{Area} \right)}{dx}=\dfrac{5}{2\sqrt{x}}-5\]
For maximum area,
\[\dfrac{d\left( \text{Area} \right)}{dx}=0\]
\[\Rightarrow \dfrac{5}{2\sqrt{x}}-5=0\]
Solving for x, we get,
\[\Rightarrow 5=5\times \left( 2\sqrt{x} \right)\]
\[\Rightarrow \sqrt{x}=\dfrac{1}{2}\]
Squaring both the sides, we get,
\[\Rightarrow x=\dfrac{1}{4}\]
So, we get the maximum area will occur at \[x=\dfrac{1}{4}.\]
Now, putting \[x=\dfrac{1}{4}\] in the value of the area as
\[\text{Area}=\dfrac{1}{2}\left[ 60+10\sqrt{x}-10x \right]\]
So,
\[\Rightarrow \text{Maximum Area}=\dfrac{1}{2}\left[ 60+10\sqrt{\dfrac{1}{4}}-10\times \dfrac{1}{4} \right]\]
Simplifying further, we get,
\[\Rightarrow \text{Maximum Area}=\dfrac{1}{2}\left[ 60+\dfrac{10}{2}-\dfrac{10}{4} \right]\]
\[\Rightarrow \text{Maximum Area}=\dfrac{1}{2}\left[ \dfrac{125}{2} \right]\]
\[\Rightarrow \text{Maximum Area}=\dfrac{125}{4}\]
Now, changing it to mixed fraction, we get,
\[\Rightarrow \text{Maximum Area}=31\dfrac{1}{4}sq.units\]
Hence, option (d) is the right answer.

Note: The derivative of \[{{x}^{n}}\] is given as \[n{{x}^{n-1}}\] and the derivative of \[\sqrt{x}\] is also found using this formula. \[\sqrt{x}\] can be written as \[{{x}^{\dfrac{1}{2}}}\] that means, we have \[n=\dfrac{1}{2}.\] So,
\[\dfrac{d\left( \sqrt{x} \right)}{dx}=\dfrac{d\left( {{x}^{\dfrac{1}{2}}} \right)}{dx}\]
Applying the same formula, we get,
\[\Rightarrow \dfrac{d\left( \sqrt{x} \right)}{dx}=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}}\]
\[\Rightarrow \dfrac{d\left( \sqrt{x} \right)}{dx}=\dfrac{1}{2}{{x}^{\dfrac{-1}{2}}}\]
As, \[{{a}^{-b}}=\dfrac{1}{{{a}^{b}}},\] so we get,
\[\Rightarrow \dfrac{d\left( \sqrt{x} \right)}{dx}=\dfrac{1}{2{{x}^{\dfrac{1}{2}}}}\]
Now, again, \[{{x}^{\dfrac{1}{2}}}=\sqrt{x},\] So,
\[\Rightarrow \dfrac{d\left( \sqrt{x} \right)}{dx}=\dfrac{1}{2\sqrt{x}}\]