Q91 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
The area bounded by the curve $y^2=4 a(x-1)$ and the lines $x=1, y=4 a$ is
- Correct answerA$\dfrac{16}{3} a^2 \text { squnits }$
- B$\dfrac{16}{3} a \text { sq units }$
- C$4 a^2 \text { sq units }$
- D$16 a^2 \text { squnits }$
Solution
The given curve is $y^2 = 4a^2(x - 1)$. This can be rewritten as $x - 1 = \dfrac{y^2}{4a^2}$, which implies $x = \dfrac{y^2}{4a^2} + 1$.
The area bounded by the curve, the line $x = 1$, and the line $y = 4a$ is the area between the curve and the y-axis (or vertical line $x=1$) from $y = 0$ to $y = 4a$.
The area $A$ is given by the integral $A = \int_{0}^{4a} (x - 1) dy$.
Substituting $x - 1 = \dfrac{y^2}{4a^2}$, we get $A = \int_{0}^{4a} \dfrac{y^2}{4a^2} dy$.
Evaluating the integral: $A = \dfrac{1}{4a^2} \left[ \dfrac{y^3}{3} \right]_{0}^{4a}$.
$A = \dfrac{1}{4a^2} \times \dfrac{(4a)^3}{3} = \dfrac{1}{4a^2} \times \dfrac{64a^3}{3}$.
$A = \dfrac{64a^3}{12a^2} = \dfrac{16}{3}a^2$.
Answer: $\dfrac{16}{3} a^2 \text { squnits }$
The area bounded by the curve, the line $x = 1$, and the line $y = 4a$ is the area between the curve and the y-axis (or vertical line $x=1$) from $y = 0$ to $y = 4a$.
The area $A$ is given by the integral $A = \int_{0}^{4a} (x - 1) dy$.
Substituting $x - 1 = \dfrac{y^2}{4a^2}$, we get $A = \int_{0}^{4a} \dfrac{y^2}{4a^2} dy$.
Evaluating the integral: $A = \dfrac{1}{4a^2} \left[ \dfrac{y^3}{3} \right]_{0}^{4a}$.
$A = \dfrac{1}{4a^2} \times \dfrac{(4a)^3}{3} = \dfrac{1}{4a^2} \times \dfrac{64a^3}{3}$.
$A = \dfrac{64a^3}{12a^2} = \dfrac{16}{3}a^2$.
Answer: $\dfrac{16}{3} a^2 \text { squnits }$
Hence, the correct answer is option: 'A'.