Area enclosed by the ellipse 2x^2 + 3y^2 = 1 | vedclass

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Question

(B) The equation of the ellipse is $2x^2 + 3y^2 = 1$.This can be rewritten in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ as:$\frac{x^2}

Vedclass · Accepted Answer

$\frac{\pi}{\sqrt{6}}$

Vedclass · Answer

(B) The equation of the ellipse is $2x^2 + 3y^2 = 1$.This can be rewritten in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ as:$\frac{x^2}{1/2} + \frac{y^2}{1/3} = 1$.Here,$a^2 = \frac{1}{2}$ and $b^2 = \frac{1}{3}$,so $a = \frac{1}{\sqrt{2}}$ and $b = \frac{1}{\sqrt{3}}$.The area of an ellipse is given by the formula $A = \pi ab$.Substituting the values,we get $A = \pi 	imes \frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{3}} = \frac{\pi}{\sqrt{6}}$ sq. units.Therefore,the correct option is $B$.

Area enclosed by the ellipse $2x^2 + 3y^2 = 1$ is . . . . . . sq. units.

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