WBJEE · Maths · Area Under Curves
Area of the figure bounded by the parabola \(y^2+8 x=16\) and \(y^2-24 x=48\) is
- A \(\frac{11}{9}\) sq. unit
- B \(\frac{32}{3} \sqrt{6}\) sq. unit
- C \(\frac{16}{3}\) sq. unit
- D \(\frac{24}{5}\) sq. unit
Answer & Solution
Correct Answer
(B) \(\frac{32}{3} \sqrt{6}\) sq. unit
Step-by-step Solution
Detailed explanation
\[ \begin{gathered} y^2=-8(x-2) \\ y^2=24(x+2) \cdots(2) \end{gathered} \] 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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