The equations of the directrices of the ellip | vedclass

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(D) The given equation of the ellipse is $16x^2 + 25y^2 = 400$.Dividing by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.Here,$a^2 = 25$ and $b^2

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(D) The given equation of the ellipse is $16x^2 + 25y^2 = 400$.Dividing by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.Here,$a^2 = 25$ and $b^2 = 16$,so $a = 5$ and $b = 4$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.The equations of the directrices for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $x = \pm \frac{a}{e}$.Substituting the values,$x = \pm \frac{5}{3/5} = \pm \frac{25}{3}$,which implies $3x = \pm 25$.Since $3x = \pm 25$ is not among the options $A, B, C$,the correct choice is $D$.

The equations of the directrices of the ellipse $16x^2 + 25y^2 = 400$ are

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