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Write an equation for an ellipse centered at the origin, which has foci at
(+-3,0) and co-vertices at
(0,+-4).

Write an equation for an ellipse centered at the origin, which has foci at $( \pm 3,0)$ and co-vertices at $(0, \pm 4)$ .

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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

( \pm 3,0)

and co-vertices at

(0, \pm 4)

.

Given Information: We are given: $\newline$ Foci: $(\pm 3, 0)$ $\newline$ Co-vertices: $(0, \pm 4)$ $\newline$ Since the foci are on the x-axis, this indicates a horizontal orientation for the ellipse.
Finding the Value of $c$ : The distance between the center and a focus is the value of $c$ in the ellipse equation. Since the foci are at $(\pm 3, 0)$ , we have: $\newline$ $c = 3$
Finding the Value of $b$ : The distance between the center and a co-vertex is the value of $b$ in the ellipse equation. Since the co-vertices are at $(0, \pm 4)$ , we have: $\newline$ $b = 4$
Standard Form of the Equation: The standard form of the equation for a horizontal ellipse centered at the origin is: $\newline$ $(x^2/a^2) + (y^2/b^2) = 1$ $\newline$ We have the value of $b$ , but we need to find the value of $a$ . The relationship between $a$ , $b$ , and $c$ for an ellipse is: $\newline$ $c^2 = a^2 - b^2$ $\newline$ We can use this to solve for $a$ .
Solving for a: Substitute the known values of $b$ and $c$ into the equation to find $a$ : $c^2 = a^2 - b^2$ $3^2 = a^2 - 4^2$ $9 = a^2 - 16$ Add $16$ to both sides to solve for $a^2$ : $a^2 = 9 + 16$ $a^2 = 25$ Take the square root of both sides to find $a$ : $a = \sqrt{25}$ $a = 5$
Substituting Values: Now that we have $a$ and $b$ , we can write the equation of the ellipse in standard form: $\newline$ $(\frac{x^2}{a^2}) + (\frac{y^2}{b^2}) = 1$ $\newline$ Substitute $a = 5$ and $b = 4$ into the equation: $\newline$ $(\frac{x^2}{5^2}) + (\frac{y^2}{4^2}) = 1$ $\newline$ Simplify the denominators: $\newline$ $(\frac{x^2}{25}) + (\frac{y^2}{16}) = 1$

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