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 bounded by the curves $y=x^2$ and $y-6=-|x|$ is

Let $A = \{(x, y) \in R^2 : y \geq 0, 2x \leq y \leq \sqrt{4-(x-1)^2}\}$ and $B = \{(x, y) \in R \times R : 0 \leq y \leq \min \{2x, \sqrt{4-(x-1)^2}\}\}$. Then the ratio of the area of $A$ to the area of $B$ is

The area (in square unit) of the region enclosed by the curves $y=x^2$ and $y=x^3$ is

The area of the shaded region is ... sq. units.

The area bounded by $y-1=-|x|$ and $y+1=|x|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module