The area bounded by the parabola {y^2} = 4ax | vedclass

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

Question

(C) The equation of the parabola is ${y^2} = 4ax$. The latus rectum is the line $x = a$.To find the area bounded by the parabola and its latus rectum,

Vedclass · Accepted Answer

$\frac{8}{3}{a^2} 	ext{ sq. unit}$

Vedclass · Answer

(C) The equation of the parabola is ${y^2} = 4ax$. The latus rectum is the line $x = a$.To find the area bounded by the parabola and its latus rectum,we integrate with respect to $x$ from $0$ to $a$.Since the parabola is symmetric about the $x$-axis,the total area is twice the area in the first quadrant.$	ext{Area} = 2 \int_0^a y \, dx = 2 \int_0^a \sqrt{4ax} \, dx$$= 2 	imes 2\sqrt{a} \int_0^a x^{1/2} \, dx$$= 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} ight]_0^a$$= 4\sqrt{a} 	imes \frac{2}{3} \left[ a^{3/2} - 0 ight]$$= \frac{8}{3} \sqrt{a} 	imes a\sqrt{a} = \frac{8}{3} a^2 	ext{ sq. unit}$

The area bounded by the parabola ${y^2} = 4ax$ and its latus rectum is:

Similar Questions

The straight line through the origin which divides the area formed by the curves $y=2x-x^2$, $y=0$, and $x=1$ into two equal halves is

The volume of the solid generated by revolving the region bounded by the parabolas $y=x^{2}$ and $x=y^{2}$ about the $y$-axis is:

The area of the region bounded by the curve $y = \sin 2x$,the $x$-axis,and the lines $x = 0$ and $x = \pi$ is . . . . . . sq. units.

Let $f(x) = x^3 - 3x^2 + 3x + 1$ and $g$ be the inverse of $f$. The area bounded by the curve $y = g(x)$ with the $x$-axis between $x = 1$ and $x = 2$ is (in square units):

Find the area bounded by the curve $y=\sin x$ between $x=0$ and $x=2 \pi$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module