The foci of the ellipse 25x^2 + 4y^2 + 100x - | vedclass

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Question

(D) Given equation: $25x^2 + 100x + 4y^2 - 4y + 100 = 0$ Complete the squares: $25(x^2 + 4x + 4) + 4(y^2 - y + 1/4) = -100 + 100 + 1$ $25(x + 2)^2 + 4

Vedclass · Accepted Answer

$\left(-2, \frac{5 \pm \sqrt{21}}{10}\right)$

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(D) Given equation: $25x^2 + 100x + 4y^2 - 4y + 100 = 0$ Complete the squares: $25(x^2 + 4x + 4) + 4(y^2 - y + 1/4) = -100 + 100 + 1$ $25(x + 2)^2 + 4(y - 1/2)^2 = 1$ Divide by $1$: $\frac{(x + 2)^2}{(1/5)^2} + \frac{(y - 1/2)^2}{(1/2)^2} = 1$ Here $a^2 = 1/25$ and $b^2 = 1/4$. Since $b > a$,the major axis is vertical $(x = -2)$. Eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1/25}{1/4}} = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$. The foci are $(h, k \pm be)$,where $(h, k) = (-2, 1/2)$. Foci $= (-2, 1/2 \pm (1/2)(\sqrt{21}/5)) = (-2, 1/2 \pm \sqrt{21}/10) = (-2, \frac{5 \pm \sqrt{21}}{10})$.

The foci of the ellipse $25x^2 + 4y^2 + 100x - 4y + 100 = 0$ are

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