Q97 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\text { The area (in sq units) enclosed by the parabola } y^2=8 x \text {, its latus-rectum and the } x \text {-axis is }$
- A$\dfrac{16 \sqrt{2}}{3}$
- B$\dfrac{32}{3}$
- Correct answerC$\dfrac{16}{3}$
- D$\dfrac{8}{3}$
Solution
The equation of the parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$, we get $4a = 8$, which implies $a = 2$.
The latus-rectum of the parabola is the line $x = a = 2$.
The area enclosed by the parabola $y^2 = 8x$, the latus-rectum $x = 2$, and the $x$-axis is given by the integral of $y$ with respect to $x$ from $x = 0$ to $x = 2$.
Since $y^2 = 8x$, we have $y = \sqrt{8x} = 2\sqrt{2}\sqrt{x}$.
Area = $\int_{0}^{2} 2\sqrt{2}\sqrt{x} dx = 2\sqrt{2} \int_{0}^{2} x^{1/2} dx$.
Evaluating the integral: $2\sqrt{2} \left[ \dfrac{x^{3/2}}{3/2} \right]_{0}^{2} = 2\sqrt{2} \times \dfrac{2}{3} \times [x^{3/2}]_{0}^{2}$.
Area = $\dfrac{4\sqrt{2}}{3} \times (2)^{3/2} = \dfrac{4\sqrt{2}}{3} \times 2\sqrt{2} = \dfrac{4 \times 2 \times 2}{3} = \dfrac{16}{3}$.
Answer: $\dfrac{16}{3}$
The latus-rectum of the parabola is the line $x = a = 2$.
The area enclosed by the parabola $y^2 = 8x$, the latus-rectum $x = 2$, and the $x$-axis is given by the integral of $y$ with respect to $x$ from $x = 0$ to $x = 2$.
Since $y^2 = 8x$, we have $y = \sqrt{8x} = 2\sqrt{2}\sqrt{x}$.
Area = $\int_{0}^{2} 2\sqrt{2}\sqrt{x} dx = 2\sqrt{2} \int_{0}^{2} x^{1/2} dx$.
Evaluating the integral: $2\sqrt{2} \left[ \dfrac{x^{3/2}}{3/2} \right]_{0}^{2} = 2\sqrt{2} \times \dfrac{2}{3} \times [x^{3/2}]_{0}^{2}$.
Area = $\dfrac{4\sqrt{2}}{3} \times (2)^{3/2} = \dfrac{4\sqrt{2}}{3} \times 2\sqrt{2} = \dfrac{4 \times 2 \times 2}{3} = \dfrac{16}{3}$.
Answer: $\dfrac{16}{3}$
Hence, the correct answer is option: 'C'.