The area of the region bounded by the parabol | vedclass

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

Question

(A) The given equation of the parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,which implies $a = 3$.The focus of the parabol

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(A) The given equation of the parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,which implies $a = 3$.The focus of the parabola is $(0, a) = (0, 3)$.The equation of the latus rectum is $y = 3$.The region is bounded by the parabola $y = \frac{x^2}{12}$ and the line $y = 3$.The points of intersection are found by substituting $y = 3$ into $x^2 = 12y$,giving $x^2 = 36$,so $x = \pm 6$.The area $A$ is given by the integral $A = \int_{-6}^{6} (3 - \frac{x^2}{12}) dx$.Since the function is even,$A = 2 \int_{0}^{6} (3 - \frac{x^2}{12}) dx$.$A = 2 [3x - \frac{x^3}{36}]_{0}^{6} = 2 [3(6) - \frac{216}{36}] = 2 [18 - 6] = 2(12) = 24$ sq. units.

The area of the region bounded by the parabola $x^2 = 12y$ and its latus rectum is . . . . . . sq. units.

Similar Questions

Let $A$ be the area bounded by the curve $y = \cos^{-1}\sqrt{1 - x^2}$,the tangent to the curve $y = \sin^{-1}x$ at $x = 0$,and the line $x = 1$. Then the value of $2(\{A\} + \text{sgn}(A))$ is (where $\{.\}$ is the fractional part function and $\text{sgn}(x)$ is the signum function).

The area of the smaller part between the circle $x^2 + y^2 = 4$ and the line $x = 1$ is

The area of the region bounded by the curves $y = \sin x$, $y = \cos x$ and the lines $x = 0$ and $x = \frac{\pi}{4}$ is

The area of the region bounded by the curves $y=x^{2}$ and $x=y^{2}$ is

The area of the region bounded by the lines $2x - y + 1 = 0$, $y = -1$, $y = 3$ and the y-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module