Area of the region bounded by the ellipse 9x^ | vedclass

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Question

(A) The given equation of the ellipse is $9x^2 + 16y^2 = 1$. This can be rewritten in the standard form $\frac{x^2}{(1/3)^2} + \frac{y^2}{(1/4)^2} = 1

Vedclass · Accepted Answer

$\frac{\pi}{48}$

Vedclass · Answer

(A) The given equation of the ellipse is $9x^2 + 16y^2 = 1$. This can be rewritten in the standard form $\frac{x^2}{(1/3)^2} + \frac{y^2}{(1/4)^2} = 1$. Here,$a = \frac{1}{3}$ and $b = \frac{1}{4}$. The total area of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\pi ab$. The area in the first quadrant is $\frac{1}{4}$ of the total area. Area $= \frac{1}{4} \pi ab = \frac{1}{4} \pi \left(\frac{1}{3}\right) \left(\frac{1}{4}\right) = \frac{\pi}{48}$ sq. units. Thus,the correct option is $A$.

Area of the region bounded by the ellipse $9x^2 + 16y^2 = 1$ in the first quadrant is . . . . . . sq. units.

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