MHT CET · Maths · Area Under Curves
The area bounded by the parabola \(y^{2}=16 x\) and its latus - rectum in the first quadrant is
- A \(128\) sq. units
- B \(\frac{64}{3}\) sq. units
- C \(\frac{128}{3}\) sq. units
- D \(64\) sq. units
Answer & Solution
Correct Answer
(B) \(\frac{64}{3}\) sq. units
Step-by-step Solution
Detailed explanation
We have parabola \(y^{2}=16 x \Rightarrow 4 a=16 \Rightarrow a=4\).
Hence coordinates of end points of latus rectum are \((4, \pm 8)\)
Required area is shaded.
\(
\begin{aligned}
\text { Area } &=4 \int_{0}^{4} \sqrt{x} \mathrm{dx}=4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4} \\
&=4 \times \frac{2}{3}\left[4^{\frac{3}{2}}-0\right]=4 \times \frac{2}{3} \times 8=\frac{64}{3} \text { sq. units }
\end{aligned}
\)
Hence coordinates of end points of latus rectum are \((4, \pm 8)\)
Required area is shaded.
\(
\begin{aligned}
\text { Area } &=4 \int_{0}^{4} \sqrt{x} \mathrm{dx}=4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4} \\
&=4 \times \frac{2}{3}\left[4^{\frac{3}{2}}-0\right]=4 \times \frac{2}{3} \times 8=\frac{64}{3} \text { sq. units }
\end{aligned}
\)
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