Write an equation for an ellipse centered at the origin, which has foci at (\pm\sqrt{8},0)(± 8 ​ ,0)(, plus minus, square root of, 8, end square root, comma, 0, )and co-vertices at (0,\pm\sqrt{10})(0,± 10 ​ )(, 0, comma, plus minus, square root of, 10, end square root, ).

Answer :

Answer:

The equation of ellipse centered at the origin

[tex]\frac{x^2}{18} +\frac{y^2}{10} =1[/tex]

Step-by-step explanation:

given the foci of ellipse (±√8,0) and c0-vertices are (0,±√10)

The foci are (-C,0) and (C ,0)

Given data (±√8,0)  

the focus has x-coordinates so the focus is  lie on x- axis.

The major axis also lie on x-axis

The minor axis lies on y-axis so c0-vertices are (0,±√10)

given focus C = ae = √8

Given co-vertices ( minor axis) (0,±b) = (0,±√10)

b= √10

The relation between the focus and semi major axes and semi minor axes are [tex]c^2=a^2-b^2[/tex]

      [tex]a^{2} = c^{2} +b^{2}[/tex]

[tex]a^{2} = (\sqrt{8} )^{2} +(\sqrt{10} )^{2}[/tex]

[tex]a^{2} =18[/tex]

[tex]a=\sqrt{18}[/tex]

The equation of ellipse formula

[tex]\frac{x^2}{a^2} +\frac{y^2}{b^2} =1[/tex]

we know that [tex]a=\sqrt{18} and b=\sqrt{10}[/tex]

Final answer:-

The equation of ellipse centered at the origin

[tex]\frac{x^2}{18} +\frac{y^2}{10} =1[/tex]

                                   

Answer: (x^2/25)+(y^2/16)=1

Step-by-step explanation:just trust me, I got the answer from khan academy and the other answer is wronggggg DO YOU UNDERSTAND?? THE OTHER ANSWER IS WRONG

Other Questions