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Area of the figure bounded by the parabola $y^2+8 x=16$ and $y^2-24 x=48$ is
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Verified Answer
The correct answer is:
$\frac{32}{3} \sqrt{6}$ sq. unit
$$
\begin{gathered}
y^2=-8(x-2) \\
y^2=24(x+2) \cdots(2)
\end{gathered}
$$
Solving (1) and $(2) \Rightarrow x=-1$
Area $=2\left[\int_{-2}^{-1} 2 \sqrt{6} \sqrt{x+2} \mathrm{dx}+\int_{-1}^2 2 \sqrt{2} \sqrt{2-x} \mathrm{dx}\right]=\frac{32}{3} \sqrt{6}$ sq. unit
\begin{gathered}
y^2=-8(x-2) \\
y^2=24(x+2) \cdots(2)
\end{gathered}
$$
Solving (1) and $(2) \Rightarrow x=-1$
Area $=2\left[\int_{-2}^{-1} 2 \sqrt{6} \sqrt{x+2} \mathrm{dx}+\int_{-1}^2 2 \sqrt{2} \sqrt{2-x} \mathrm{dx}\right]=\frac{32}{3} \sqrt{6}$ sq. unit
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