Respuesta :
Answer:
The equation of ellipse is
[tex]\frac{x^2}{41}+\frac{y^2}{16}=1[/tex]
Step-by-step explanation:
The equation of an ellipse is
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]
where(h,k) is the center and c is distance from the center to the foci is given by [tex]a^2-b^2=c^2[/tex]. 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 [tex](\pm\sqrt{41},0)[/tex] is
=[tex](\frac{\sqrt{41}+(-\sqr{41})}2,\frac{0+0}2)[/tex]
=(0,0)
a is the distance between the center and the vertices.
So, [tex]a=\sqrt{(0-\sqrt{41})^2+(0-0)^2}[/tex]
[tex]=\sqrt{41}[/tex]
c is the distance between the center and the foci.
So, [tex]c=\sqrt{(0-5)^2+(0-0)^2[/tex]
=5
[tex]a^2-b^2=c^2[/tex]
[tex]\Rightarrow (\sqrt41)^2-b^2=5^2[/tex]
[tex]\Rightarrow b^2=(\sqrt41)^2- 5^2[/tex]
[tex]\Rightarrow b^2=41-25[/tex]
[tex]\Rightarrow b^2=16[/tex]
The equation of ellipse is
[tex]\frac{(x-0)^2}{(\sqrt{41})^2}+\frac{(y-0)^2}{16}=1[/tex]
[tex]\Rightarrow \frac{x^2}{41}+\frac{y^2}{16}=1[/tex]
