How do you find the equation in standard form of an ellipse that passes through the given points: (-8, 0), (8, 0), (0, -4), (0, 4)?

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Answer 1 · 2016-12-21T14:49:07

The equation is $\frac{x^{2}}{64} + \frac{y^{2}}{16} = 1$ $x^2/64+y^2/16=1$

We use the equation the standard ellipse as the center is $(0, 0)$ $(0,0)$

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $x^2/a^2+y^2/b^2=1$

Let's take the vertex $(8, 0)$ $(8,0)$ , then

$\frac{64}{a^{2}} + 0 = 1$ $64/a^2+0=1$ , $\Rightarrow$ $=>$ , $a = 8$ $a=8$

Let's take the vertex $(0, 4)$ $(0,4)$ , then

$0 + \frac{16}{b^{2}} = 1$ $0+16/b^2=1$ , $\Rightarrow$ $=>$ , $b = 4$ $b=4$

The equation of the ellipse is

$\frac{x^{2}}{64} + \frac{y^{2}}{16} = 1$ $x^2/64+y^2/16=1$

graph{x^2/64+y^2/16=1 [-12.66, 12.65, -6.33, 6.33]}