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Write an equation for an ellipse centered at the origin, which has foci at
(0,+-15) and vertices at
(0,+-25).

Write an equation for an ellipse centered at the origin, which has foci at $(0, \pm 15)$ and vertices at $(0, \pm 25)$ .

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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 15)

and vertices at

(0, \pm 25)

.

Identify coordinates of foci and vertices: Identify the coordinates of the foci and vertices. $\newline$ Foci: $(0, \pm 15)$ $\newline$ Vertices: $(0, \pm 25)$ $\newline$ Since both foci and vertices lie on the y-axis, the ellipse is vertical.
Determine values of $a$ and $c$ : Determine the values of $a$ and $c$ . The distance from the center to a vertex is ' $a$ ', and the distance from the center to a focus is ' $c$ '. $a = 25$ (since the vertices are at $(0, \pm 25)$ ) $c = 15$ (since the foci are at $(0, \pm 15)$ )
Calculate value of b: Calculate the value of b using the relationship $c^2 = a^2 - b^2$ . $\newline$ $c^2 = a^2 - b^2$ $\newline$ $15^2 = 25^2 - b^2$ $\newline$ $225 = 625 - b^2$ $\newline$ $b^2 = 625 - 225$ $\newline$ $b^2 = 400$ $\newline$ $b = \sqrt{400}$ $\newline$ $b = 20$
Write equation in standard form: Write the equation of the ellipse in standard form. $\newline$ Since the ellipse is vertical, the standard form of the equation is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ . $\newline$ Substitute the values of a and b: $\newline$ $\frac{x^2}{20^2} + \frac{y^2}{25^2} = 1$ $\newline$ $\frac{x^2}{400} + \frac{y^2}{625} = 1$
Simplify equation if necessary: Simplify the equation if necessary. $\newline$ The equation is already in its simplest form, so no further simplification is needed. $\newline$ Final equation: $\frac{x^2}{400} + \frac{y^2}{625} = 1$

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