The eccentricity of the ellipse 25x^2 + 16y^2 | vedclass

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(D) Given equation: $25x^2 + 16y^2 - 150x - 175 = 0$Complete the square for $x$: $25(x^2 - 6x) + 16y^2 = 175$$25(x^2 - 6x + 9) + 16y^2 = 175 + 225$$25

Vedclass · Accepted Answer

$3/5$

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(D) Given equation: $25x^2 + 16y^2 - 150x - 175 = 0$Complete the square for $x$: $25(x^2 - 6x) + 16y^2 = 175$$25(x^2 - 6x + 9) + 16y^2 = 175 + 225$$25(x - 3)^2 + 16y^2 = 400$Divide by $400$: $\frac{(x - 3)^2}{16} + \frac{y^2}{25} = 1$Here,$a^2 = 25$ and $b^2 = 16$. Since $a^2 > b^2$,the ellipse is vertical.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 eccentricity of the ellipse $25x^2 + 16y^2 - 150x - 175 = 0$ is

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