If the area of the ellipse frac{x^{2}}{25}+fr | vedclass

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(A) The area of an ellipse given by the equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is calculated as $\pi |ab|$.Given the equation $\frac{x^{

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$\pm 4$

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(A) The area of an ellipse given by the equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is calculated as $\pi |ab|$.Given the equation $\frac{x^{2}}{25}+\frac{y^{2}}{\lambda^{2}}=1$,we identify $a^{2} = 25$ and $b^{2} = \lambda^{2}$.Thus,$a = 5$ and $b = |\lambda|$.The area is given as $20 \pi$ sq units.Substituting these values into the area formula: $\pi 	imes 5 	imes |\lambda| = 20 \pi$.Dividing both sides by $5 \pi$,we get $|\lambda| = 4$.Therefore,$\lambda = \pm 4$.

If the area of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{\lambda^{2}}=1$ is $20 \pi$ sq units,then $\lambda$ is

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