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

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,
$\Rightarrow \dfrac{d\left( \sqrt{x} \right)}{dx}=\dfrac{1}{2\sqrt{x}}$