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Write an equation for an ellipse centered at the origin, which has foci at
(0,+-6) and vertices at
(0,+-10).

Write an equation for an ellipse centered at the origin, which has foci at $(0, \pm 6)$ and vertices at $(0, \pm 10)$ .

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Write equations of ellipses in standard form using properties

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Q. Write an equation for an ellipse centered at the origin, which has foci at

(0, \pm 6)

and vertices at

(0, \pm 10)

.

Given Information: We are given: $\newline$ Foci: $(0, \pm 6)$ $\newline$ Vertices: $(0, \pm 10)$ $\newline$ Since the vertices and foci are on the y-axis, this indicates a vertical orientation for the ellipse.
Center of the Ellipse: The center $(h, k)$ of the ellipse is at the origin, so $h = 0$ and $k = 0$ .
Value of 'a': The distance from the center to a vertex is the value of 'a'. Since the vertices are at $(0, \pm 10)$ , we have: $\newline$ $a = 10$
Value of 'c': The distance from the center to a focus is the value of 'c'. Since the foci are at $(0, \pm 6)$ , we have: $\newline$ $c = 6$
Calculation of 'b': We need to find the value of 'b'. The relationship between $a$ , $b$ , and $c$ for an ellipse is $c^2 = a^2 - b^2$ . We can solve for $b^2$ :
$b^2 = a^2 - c^2$
$b^2 = 10^2 - 6^2$
$b^2 = 100 - 36$
$b^2 = 64$
Value of 'b': Now we can find the value of 'b': $\newline$ $b = \sqrt{b^2}$ $\newline$ $b = \sqrt{64}$ $\newline$ $b = 8$
Equation of the Ellipse: We can now write the equation of the ellipse in standard form. Since the ellipse is vertically oriented, the equation is: $\newline$ $(x - h)^2/b^2 + (y - k)^2/a^2 = 1$ $\newline$ Substituting the values of $h$ , $k$ , $a$ , and $b$ , we get: $\newline$ $(x - 0)^2/8^2 + (y - 0)^2/10^2 = 1$ $\newline$ $x^2/64 + y^2/100 = 1$

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