The area enclosed by the ellipse 25x^2 + 16y^ | vedclass

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(B) The given equation of the ellipse is $25x^2 + 16y^2 = 400$. Dividing both sides by $400$,we get: $\frac{25x^2}{400} + \frac{16y^2}{400} = 1$ $\fra

Vedclass · Accepted Answer

$20$

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(B) The given equation of the ellipse is $25x^2 + 16y^2 = 400$. Dividing both sides by $400$,we get: $\frac{25x^2}{400} + \frac{16y^2}{400} = 1$ $\frac{x^2}{16} + \frac{y^2}{25} = 1$ Comparing this with the standard equation $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,we have $a^2 = 25$ and $b^2 = 16$. Thus,$a = 5$ and $b = 4$. The area of an ellipse is given by the formula $A = \pi ab$. Substituting the values,$A = \pi 	imes 5 	imes 4 = 20\pi$ sq. units.

The area enclosed by the ellipse $25x^2 + 16y^2 = 400$ is . . . . . . sq. units. (in $\pi$)

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