Question

Write an equation for an ellipse centered at the origin, which has foci at (\pm5,0)(±5,0)(, plus minus, 5, comma, 0, )and vertices at (\pm\sqrt{41},0)(± 41 ​ ,0)(, plus minus, square root of, 41, end square root, comma, 0, ).

Answer 1

Answer:

The equation of ellipse is

$\frac{x^2}{41}+\frac{y^2}{16}=1$

Step-by-step explanation:

The equation of an ellipse is

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

where(h,k) is the center and c is distance from the center to the foci is given by $a^2-b^2=c^2$. a is the distance from the center to the vertices and b is the distance from the center to the co-vertices.

The center of the ellipse is the mid-point of the vertices.

The mid point of the vertices $(\pm\sqrt{41},0)$ is

=$(\frac{\sqrt{41}+(-\sqr{41})}2,\frac{0+0}2)$

=(0,0)

a is the distance between the center and the vertices.

So, $a=\sqrt{(0-\sqrt{41})^2+(0-0)^2}$

        $=\sqrt{41}$

c is the distance between the center and the foci.

So, $c=\sqrt{(0-5)^2+(0-0)^2$

         =5

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

$\Rightarrow (\sqrt41)^2-b^2=5^2$

$\Rightarrow b^2=(\sqrt41)^2- 5^2$

$\Rightarrow b^2=41-25$

$\Rightarrow b^2=16$

The equation of ellipse is

$\frac{(x-0)^2}{(\sqrt{41})^2}+\frac{(y-0)^2}{16}=1$

$\Rightarrow \frac{x^2}{41}+\frac{y^2}{16}=1$