How do you write an equation of an ellipse in standard form given foci (+/-4,0) and co-vertices at (0,+/-2)?

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Answer 1 · 2016-01-29T08:54:06

$\frac{x^{2}}{20} + \frac{y^{2}}{4} = 1$ $x^2/20+y^2/4=1$

From the given data:

Foci(4, 0) and (-4, 0)
Co-vertices (0, 2) and (0, -2) are the endpoints of the minor axis.

The Center is at the Origin $(0, 0)$ $(0, 0)$ by inspection.

The standard form equation for Horizontal major axis Ellipse

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x-h)^2/a^2+(y-k)^2/b^2=1$

$b = 2$ $b=2$ the length of 1/2 of the minor axis called semi-minor axis.
$c = 4$ $c=4$ the distance from center to a focus.

$a$ $a$ is the length of 1/2 of the major axis also called semi-major axis.

Compute for $a$ $a$

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

$a = \sqrt{c^{2} + b^{2}}$ $a=sqrt(c^2+b^2)$

$a = \sqrt{4^{2} + 2^{2}}$ $a=sqrt(4^2+2^2)$

$a = \sqrt{16 + 4}$ $a=sqrt(16+4)$

$a = \sqrt{20} = 2 \sqrt{5}$ $a=sqrt(20)=2sqrt(5)$

Use now the equation

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x-h)^2/a^2+(y-k)^2/b^2=1$ with Center $(h, k) = (0, 0)$ $(h, k)=(0, 0)$

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

which simplifies to

$\frac{x^{2}}{20} + \frac{y^{2}}{4} = 1$ $x^2/20+y^2/4=1$

Kindly check the graph...

graph{(x^2/20+y^2/4-1)=0[-10,10,-5,5]}

Have a nice day !!! from the Philippines .