The area included between the parabola y=frac | vedclass

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

Question

(D) Given curves are $y=\frac{x^2}{4 a}$ and $y=\frac{8 a^3}{x^2+4 a^2}$.For the point of intersection,equate the two expressions:$\frac{x^2}{4 a} = \

Vedclass · Accepted Answer

$a^2(2 \pi-\frac{4}{3})$

Vedclass · Answer

(D) Given curves are $y=\frac{x^2}{4 a}$ and $y=\frac{8 a^3}{x^2+4 a^2}$.For the point of intersection,equate the two expressions:$\frac{x^2}{4 a} = \frac{8 a^3}{x^2+4 a^2}$$x^2(x^2+4 a^2) = 32 a^4$$x^4+4 a^2 x^2 - 32 a^4 = 0$$(x^2+8 a^2)(x^2-4 a^2) = 0$Since $x^2 = -8 a^2$ is not possible,we have $x^2 = 4 a^2$,so $x = \pm 2 a$.The area $A$ is given by $2 \int_0^{2 a} (\frac{8 a^3}{x^2+4 a^2} - \frac{x^2}{4 a}) dx$.$A = 2 [8 a^3 \int_0^{2 a} \frac{1}{x^2+(2 a)^2} dx - \frac{1}{4 a} \int_0^{2 a} x^2 dx]$$A = 2 [8 a^3 \cdot \frac{1}{2 a} 	an^{-1}(\frac{x}{2 a}) |_0^{2 a} - \frac{1}{4 a} \cdot \frac{x^3}{3} |_0^{2 a}]$$A = 2 [4 a^2 	an^{-1}(1) - \frac{1}{4 a} \cdot \frac{8 a^3}{3}]$$A = 2 [4 a^2 \cdot \frac{\pi}{4} - \frac{2 a^2}{3}]$$A = 2 a^2 (\pi - \frac{2}{3}) = a^2(2 \pi - \frac{4}{3})$.

The area included between the parabola $y=\frac{x^2}{4 a}$ and the curve $y=\frac{8 a^3}{x^2+4 a^2}$ is

Similar Questions

The area enclosed between the curves $y^2 = 4x$ and $y = |x|$ is

Let $A$ be the area of the region $\{(x, y): y \geq x^2, y \geq(1-x)^2, y \leq 2x(1-x)\}$. Then $540A$ is equal to

The area of the smaller portion between the curves $x^2 + y^2 = 8$ and $y^2 = 2x$ is

Let the area of the region $\{(x, y): 2y \leq x^2+3, y +|x| \leq 3, y \geq|x-1|\}$ be $A$. Then $6A$ is equal to:

One of the points of intersection of the curves $y=1+3x-2x^2$ and $y=\frac{1}{x}$ is $\left(\frac{1}{2}, 2\right)$. Let the area of the region enclosed by these curves be $\frac{1}{24}(\ell \sqrt{5}+m)-n \log_{e}(1+\sqrt{5})$,where $\ell, m, n \in N$. Then $\ell+m+n$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module