The area (in square units) of the region boun | vedclass

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

Question

(B) Given equations are $y^2=4(x-2)$ and $y=2x-8$. From the line equation,$2x = y+8$,so $x = \frac{y+8}{2} = \frac{y}{2} + 4$. Substituting $x$ in the

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Given equations are $y^2=4(x-2)$ and $y=2x-8$. From the line equation,$2x = y+8$,so $x = \frac{y+8}{2} = \frac{y}{2} + 4$. Substituting $x$ in the parabola equation: $y^2 = 4(\frac{y}{2} + 4 - 2) = 4(\frac{y}{2} + 2) = 2y + 8$. $y^2 - 2y - 8 = 0 \implies (y-4)(y+2) = 0$. Thus,the intersection points are $y=4$ and $y=-2$. The area $A$ is given by $\int_{-2}^{4} (x_{line} - x_{parabola}) dy$. $x_{line} = \frac{y+8}{2}$ and $x_{parabola} = \frac{y^2}{4} + 2$. $A = \int_{-2}^{4} (\frac{y+8}{2} - (\frac{y^2}{4} + 2)) dy = \int_{-2}^{4} (\frac{y}{2} + 4 - \frac{y^2}{4} - 2) dy = \int_{-2}^{4} (\frac{y}{2} + 2 - \frac{y^2}{4}) dy$. $A = [\frac{y^2}{4} + 2y - \frac{y^3}{12}]_{-2}^{4}$. $A = (\frac{16}{4} + 8 - \frac{64}{12}) - (\frac{4}{4} - 4 - \frac{-8}{12}) = (4 + 8 - \frac{16}{3}) - (1 - 4 + \frac{2}{3}) = (12 - \frac{16}{3}) - (-3 + \frac{2}{3}) = \frac{20}{3} - (-\frac{7}{3}) = \frac{27}{3} = 9$ square units.

The area (in square units) of the region bounded by the parabola $y^2=4(x-2)$ and the line $y=2x-8$ is:

Similar Questions

The area (in square units) of the region bounded by the curves $y + 2x^2 = 0$ and $y + 3x^2 = 1$ is equal to

The area (in sq. units) enclosed by the curves $y=2x-x^2$ and $y=x^2-2x-6$ is

The area bounded by $y=x+1$, $y=\cos x$ and the $X$-axis is

Let the area enclosed by the lines $x + y = 2, y = 0, x = 0$ and the curve $f(x) = \min \left\{x^2 + \frac{3}{4}, 1 + [x]\right\}$,where $[x]$ denotes the greatest integer $\leq x$,be $A$. Then the value of $12A$ is $............$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module