What is the foci of $25 x^{2} + 36 y^{2} = 900$ $25x^2+36y^2=900$ ?

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Answer 1 · 2016-10-31T19:12:28

$(- \sqrt{11}, 0) and (\sqrt{11}, 0)$ $(-sqrt(11),0) and (sqrt(11),0)$

Standard form is:

$\frac{{(x - 0)}^{2}}{6^{2}} + \frac{{(y - 0)}^{2}}{5^{2}} = 1$ $(x - 0)^2/6^2 + (y - 0)^2/5^2 = 1$

The focal distance, c, is computed using

$c^{2} = a^{2} - b^{2}$ $c^2 = a^2 - b^2$

$c^{2} = 36 - 25$ $c^2 = 36 - 25$

$c^{2} = 11$ $c^2 = 11$

$c = \sqrt{11}$ $c = sqrt(11)$

What is the foci of 25x2+36y2=90025x^2+36y^2=900?