In the figure,AOBA is the part of the ellipse | vedclass

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

Question

(A) The given equation of the ellipse is $9x^{2} + y^{2} = 36$,which can be written as $\frac{x^{2}}{4} + \frac{y^{2}}{36} = 1$ or $\frac{x^{2}}{2^{2}

Vedclass · Accepted Answer

$3\pi - 6$

Vedclass · Answer

(A) The given equation of the ellipse is $9x^{2} + y^{2} = 36$,which can be written as $\frac{x^{2}}{4} + \frac{y^{2}}{36} = 1$ or $\frac{x^{2}}{2^{2}} + \frac{y^{2}}{6^{2}} = 1$.From the equation,$y = \sqrt{36 - 9x^{2}} = 3\sqrt{4 - x^{2}}$.The equation of the chord $AB$ passing through $A(2, 0)$ and $B(0, 6)$ is given by $\frac{x}{2} + \frac{y}{6} = 1$,which simplifies to $y = 6 - 3x$.The area of the shaded region is the area under the ellipse minus the area under the line $AB$ from $x = 0$ to $x = 2$.Area $= \int_{0}^{2} (y_{	ext{ellipse}} - y_{	ext{line}}) dx = \int_{0}^{2} (3\sqrt{4 - x^{2}} - (6 - 3x)) dx$.$= 3 \int_{0}^{2} \sqrt{2^{2} - x^{2}} dx - \int_{0}^{2} (6 - 3x) dx$.Using the formula $\int \sqrt{a^{2} - x^{2}} dx = \frac{x}{2}\sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{x}{a})$:$= 3 \left[ \frac{x}{2}\sqrt{4 - x^{2}} + \frac{4}{2}\sin^{-1}(\frac{x}{2}) ight]_{0}^{2} - \left[ 6x - \frac{3x^{2}}{2} ight]_{0}^{2}$.$= 3 \left[ (0 + 2\sin^{-1}(1)) - (0 + 0) ight] - \left[ (12 - 6) - (0 - 0) ight]$.$= 3 \left[ 2 	imes \frac{\pi}{2} ight] - 6 = 3\pi - 6$ square units.

In the figure,$AOBA$ is the part of the ellipse $9x^{2} + y^{2} = 36$ in the first quadrant such that $OA = 2$ and $OB = 6$. Find the area between the arc $AB$ and the chord $AB$.

Similar Questions

The area of the region $\{(x, y) : y \le x - |x|, y \le |x \sin x|, y \ge 0\}$ is:

Area of the region bounded by the curve $y = \cos x$,$x = -\frac{\pi}{2}$ and $x = \pi$ is . . . . . . sq. units.

If the tangent to the curve $y = 1 - x^2$ at $x = \alpha,$ where $0 < \alpha < 1,$ meets the axes at $P$ and $Q.$ Also $\alpha$ varies,the minimum value of the area of the triangle $OPQ$ is $k$ times the area bounded by the axes and the part of the curve for which $0 < x < 1,$ then $k$ is equal to

The area bounded by the $x$-axis and the curve $y = \sin x$ between $x = 0$ and $x = \pi$ is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module