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Write an equation for an ellipse centered at the origin, which has foci at
(0,+-sqrt62) and co-vertices at
(+-5,0).

Write an equation for an ellipse centered at the origin, which has foci at $(0, \pm \sqrt{62})$ and co-vertices at $( \pm 5,0)$ .

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Write equations of ellipses in standard form using properties

Full solution

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

(0, \pm \sqrt{62})

and co-vertices at

( \pm 5,0)

.

Given Information: We are given: $\newline$ Foci: $(0, \pm\sqrt{62})$ $\newline$ Co-vertices: $(\pm5, 0)$ $\newline$ Since the foci are on the y-axis, this indicates a vertical orientation for the ellipse.
Finding $c$ : The distance between the center and a focus is the value of $c$ in the ellipse equation. We can find $c$ by using the coordinates of the foci. $c = \sqrt{62}$
Finding $b$ : The distance between the center and a co-vertex is the value of $b$ in the ellipse equation. We can find $b$ by using the coordinates of the co-vertices. $\newline$ $b = 5$
Finding $a$ : To find the value of $a$ , we use the relationship between $a$ , $b$ , and $c$ in an ellipse, which is $c^2 = a^2 - b^2$ . We already know $c$ and $b$ , so we can solve for $a$ . $c^2 = a^2 - b^2$ $(\sqrt{62})^2 = a^2 - 5^2$ $62 = a^2 - 25$ $a^2 = 62 + 25$ $a^2 = 87$ $a = \sqrt{87}$
Writing the Equation: Now we have all the values needed to write the equation of the ellipse in standard form. Since the ellipse is vertically oriented, the $a^2$ term will be under the $y^2$ term and the $b^2$ term will be under the $x^2$ term. $\newline$ The standard form of the ellipse is: $\newline$ $(x - h)^2/b^2 + (y - k)^2/a^2 = 1$ $\newline$ Plugging in the values for $h$ , $k$ , $a$ , and $b$ , we get: $\newline$ $(x - 0)^2/5^2 + (y - 0)^2/\sqrt{87}^2 = 1$ $\newline$ $x^2/25 + y^2/87 = 1$

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