The area (in sq units) between the curve y^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$. The latus rectum of the parabola is the line $x = a

Vedclass · Accepted Answer

$\frac{32}{3}$

Vedclass · Answer

(A) The given parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,so $a = 2$. The latus rectum of the parabola is the line $x = a = 2$. The area bounded by the curve and the latus rectum is symmetric about the $x$-axis. The required area is $A = 2 \int_0^2 y \, dx = 2 \int_0^2 \sqrt{8x} \, dx$. $A = 2 	imes 2\sqrt{2} \int_0^2 x^{1/2} \, dx = 4\sqrt{2} \left[ \frac{x^{3/2}}{3/2} ight]_0^2$. $A = 4\sqrt{2} 	imes \frac{2}{3} 	imes (2)^{3/2} = \frac{8\sqrt{2}}{3} 	imes 2\sqrt{2} = \frac{16 	imes 2}{3} = \frac{32}{3}$ sq units.

The area (in sq units) between the curve $y^2 = 8x$ and its latus rectum is

Similar Questions

The area of the region lying in the first quadrant bounded by $y=4x^2$,$x=0$,$y=2$,and $y=4$ is

If the area of the region $\{(x, y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1\}$ is $(\log_e 2) - \frac{1}{7}$,then the value of $7a - 3$ is equal to:

The area bounded by the $x$-axis and the curve $y = \sin x$ between $x = 0$ and $x = \pi$ is:

The area bounded by the curve $y = x^3$,the $x$-axis,and the ordinates $x = 1$ and $x = 4$ is:

The area of the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ and its latus rectum is . . . . . . sq. units. (where $e$ denotes the eccentricity of the ellipse).

Vedclass Test Series

Exam Paper Generator

Online Exam Module