Question

Write an equation for an ellipse centered at the origin, which has foci at (\pm\sqrt{12},0)(± 12 ​ ,0)left parenthesis, plus minus, square root of, 12, end square root, comma, 0, right parenthesis and vertices at (\pm\sqrt{37},0)(± 37 ​ ,0)left parenthesis, plus minus, square root of, 37, end square root, comma, 0, right parenthesis

Answer 1

Answer:

$\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}$

Step-by-step explanation:

Given that :

the foci of the ellipse is (±√12,0) and C0-vertices are (0,±√37)

The foci are (-C,0) and (C ,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,±√37)

The given focus C = ae = √12

Given co-vertices ( minor axis) (0,±b) = (0,±√37)

b= √37

We can therefore express the  relation between the focus and semi major axes and semi minor axes as:

$\mathbf{c^2 = a^2 - b^2 } \\ \\ \mathbf{a^2 = c^2 + b^2 } \\ \\ \mathbf{c^2 = ( \sqrt12)^2 - (\sqrt 37)^2 } \\ \\ \mathbf{c^2 = 49 } \\ \\ \mathbf{c = \sqrt{49 }}$

The equation of ellipse formula is:

$\dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1$

and we know that $\mathbf{a=\sqrt{49} \ \ and \ \ b=\sqrt{37}}$

Thus ; the equation of the ellipse at the origin is

$\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}$