Area enclosed by the ellipse 9x^2 + 4y^2 = 1 | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{\pi}{24}$

Vedclass · Answer

(C) The given equation of the ellipse is $9x^2 + 4y^2 = 1$. This can be rewritten in the standard form $\frac{x^2}{(1/3)^2} + \frac{y^2}{(1/2)^2} = 1$. Here,$a = \frac{1}{3}$ and $b = \frac{1}{2}$. The area of the entire ellipse is given by the formula $A = \pi ab$. $A = \pi 	imes \frac{1}{3} 	imes \frac{1}{2} = \frac{\pi}{6}$. Since the ellipse is symmetric about both axes,the area in the first quadrant is one-fourth of the total area. Area in the first quadrant $= \frac{1}{4} 	imes \frac{\pi}{6} = \frac{\pi}{24}$. Thus,the correct option is $C$.

Area enclosed by the ellipse $9x^2 + 4y^2 = 1$ in the first quadrant is . . . . . . .

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