MHT CET · Maths · Area Under Curves
The area of the region bounded by the parabola \(y^{2}=8 x\) and its latus rectum is
- A \(\frac{16}{3}\) sq. units
- B \(\frac{8}{3}\) sq. units
- C \(\frac{32}{3}\) sq. units
- D \(\frac{4}{3}\) sq. units
Answer & Solution
Correct Answer
(C) \(\frac{32}{3}\) sq. units
Step-by-step Solution
Detailed explanation
We have parabola \(\mathrm{y}^{2}=8 \mathrm{x} \Rightarrow 4 \mathrm{a}=8 \Rightarrow \mathrm{a}=2\)
Hence coordinates of latus rectum are
\(=(\mathrm{a}, \pm 2 \mathrm{a}) \text { i.e. }(2,4) \text { and }(2,-4)\)
Required area is shaded in figure.
\(\begin{aligned}
\therefore A &=2 \int_{0}^{2} \sqrt{8 x} d x=4 \sqrt{2} \int_{0}^{2} x^{\frac{1}{2}} d x \\
&=4 \sqrt{2}\left[\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_{0}^{2}=\frac{8 \sqrt{2}}{3}(2 \sqrt{2})=\frac{32}{3}
\end{aligned}\)
Hence coordinates of latus rectum are
\(=(\mathrm{a}, \pm 2 \mathrm{a}) \text { i.e. }(2,4) \text { and }(2,-4)\)
Required area is shaded in figure.
\(\begin{aligned}
\therefore A &=2 \int_{0}^{2} \sqrt{8 x} d x=4 \sqrt{2} \int_{0}^{2} x^{\frac{1}{2}} d x \\
&=4 \sqrt{2}\left[\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_{0}^{2}=\frac{8 \sqrt{2}}{3}(2 \sqrt{2})=\frac{32}{3}
\end{aligned}\)
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