Question

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 1

The equation of the ellipse centered at the origin with foci at (±√8, 0) and co-vertices at (0, ±√10) is x²/18 + y²/10 = 1.

Step-by-step explanation:

The equation of an ellipse with the center at the origin, foci at (±√8, 0), and co-vertices at (0, ±√10) can be derived using the standard form of an ellipse's equation. The distance between the center and a focus, denoted by 'c', is √8 in this case. The distance from the center to the co-vertices, 'b', is √10. As the foci lie along the x-axis, the major axis is horizontal and the longer dimension 'a' must satisfy the relationship a2 = b2 + c2. Plugging the given values into this relationship gives a2 = 10 + 8 = 18. Thus, the equation of the ellipse is x2/18 + y2/10 = 1.

Answer 2

The equation of ellipse centered at the origin

`(x^2)/(18) +(y^2)/(10) =1`

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
`c^2=a^2-b^2`

`a^(2) = c^(2) +b^(2)`

`a^(2) = (√(8) )^(2) +(√(10) )^(2)`

`a^(2) =18`

`a=√(18)`

The equation of ellipse formula

`(x^2)/(a^2) +(y^2)/(b^2) =1`

we know that
`a=√(18) and b=√(10)`

Final answer:-

The equation of ellipse centered at the origin

`(x^2)/(18) +(y^2)/(10) =1`