# Area enclosed by the curve \[\pi \left[ 4{{\left( x-\sqrt{2} \right)}^{2}}+{{y}^{2}} \right]=8\] is

(A) \[\pi \]sq units

(B) \[2\]sq units

(C) \[3\pi \]sq units

(D) \[4\]sq units

Answer

Verified

363.9k+ views

Hint: First try to make a rough sketch. Now, break the entire area into parts as per the line of symmetry. Then find the area under one curve and multiply by four to get the total area. Thus, break the integral up into two smaller intervals by looking at the diagram and proceed. Consider the given equation of the curve.

\[\pi \left[ 4{{\left( x-\sqrt{2} \right)}^{2}}+{{y}^{2}} \right]=8\]

Dividing both sides by 8, we get

\[\dfrac{4\pi {{\left( x-\sqrt{2} \right)}^{2}}}{8}+\dfrac{\pi {{y}^{2}}}{8}=1\]

The above equation can also be written as

\[\dfrac{{{\left( x-\sqrt{2} \right)}^{2}}}{\dfrac{8}{4\pi }}+\dfrac{{{y}^{2}}}{\dfrac{8}{\pi }}=1\]

\[\dfrac{{{\left( x-\sqrt{2} \right)}^{2}}}{\dfrac{2}{\pi }}+\dfrac{{{y}^{2}}}{\dfrac{8}{\pi }}=1..........(i)\]

The above equation resembles an ellipse.

The general equation of ellipse is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ . Comparing the equation (i) with the general equation of ellipse, we get

\[a=\sqrt{\dfrac{2}{\pi }}\]and \[b=\sqrt{\dfrac{8}{\pi }}\]

And the centre of the ellipse is $\left( \sqrt{2},0 \right)$ .

We know the area of the ellipse is given by the formula,

\[\text{Area=}\pi ab\]

Substituting the value of $a,b$, we get

\[\begin{align}

& \Rightarrow \text{Area =}\pi \times \sqrt{\dfrac{2}{\pi }}\times \sqrt{\dfrac{8}{\pi }} \\

& \Rightarrow \text{Area =}\dfrac{\pi }{\pi }\times \sqrt{2\times 8} \\

& \Rightarrow \text{Area =}\sqrt{16}=4 \\

\end{align}\]

So the area enclosed by the curve \[\pi \left[ 4{{\left( x-\sqrt{2} \right)}^{2}}+{{y}^{2}} \right]=8\] is $4$ units.

Hence the option (D) is the correct answer.

Note: The probability of mistake is that student might take \[a=\dfrac{8}{4\pi }\] and \[b=\dfrac{8}{\pi }\]instead of \[a=\sqrt{\dfrac{8}{4\pi }}\] and \[b=\sqrt{\dfrac{8}{\pi }}\] as the general equation of ellipse is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.

And using integration to find the area of the ellipse will result in a complex number.

\[\pi \left[ 4{{\left( x-\sqrt{2} \right)}^{2}}+{{y}^{2}} \right]=8\]

Dividing both sides by 8, we get

\[\dfrac{4\pi {{\left( x-\sqrt{2} \right)}^{2}}}{8}+\dfrac{\pi {{y}^{2}}}{8}=1\]

The above equation can also be written as

\[\dfrac{{{\left( x-\sqrt{2} \right)}^{2}}}{\dfrac{8}{4\pi }}+\dfrac{{{y}^{2}}}{\dfrac{8}{\pi }}=1\]

\[\dfrac{{{\left( x-\sqrt{2} \right)}^{2}}}{\dfrac{2}{\pi }}+\dfrac{{{y}^{2}}}{\dfrac{8}{\pi }}=1..........(i)\]

The above equation resembles an ellipse.

The general equation of ellipse is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ . Comparing the equation (i) with the general equation of ellipse, we get

\[a=\sqrt{\dfrac{2}{\pi }}\]and \[b=\sqrt{\dfrac{8}{\pi }}\]

And the centre of the ellipse is $\left( \sqrt{2},0 \right)$ .

We know the area of the ellipse is given by the formula,

\[\text{Area=}\pi ab\]

Substituting the value of $a,b$, we get

\[\begin{align}

& \Rightarrow \text{Area =}\pi \times \sqrt{\dfrac{2}{\pi }}\times \sqrt{\dfrac{8}{\pi }} \\

& \Rightarrow \text{Area =}\dfrac{\pi }{\pi }\times \sqrt{2\times 8} \\

& \Rightarrow \text{Area =}\sqrt{16}=4 \\

\end{align}\]

So the area enclosed by the curve \[\pi \left[ 4{{\left( x-\sqrt{2} \right)}^{2}}+{{y}^{2}} \right]=8\] is $4$ units.

Hence the option (D) is the correct answer.

Note: The probability of mistake is that student might take \[a=\dfrac{8}{4\pi }\] and \[b=\dfrac{8}{\pi }\]instead of \[a=\sqrt{\dfrac{8}{4\pi }}\] and \[b=\sqrt{\dfrac{8}{\pi }}\] as the general equation of ellipse is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.

And using integration to find the area of the ellipse will result in a complex number.

Last updated date: 28th Sep 2023

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