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

Given Information: We are given: Foci: (0, ±√62) Co-vertices: (±5, 0) 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 = √62Finding 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. b = 5Finding 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 (√62)^2 = a^2 - 5^2 62 = a^2 - 25 a^2 = 62 + 25 a^2 = 87 a = √87Writing 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. The standard form of the ellipse is: (x - h)^2/b^2 + (y - k)^2/a^2 = 1 Plugging in the values for h, k, a, and b, we get: (x - 0)^2/5^2 + (y - 0)^2/√87^2 = 1 x^2/25 + y^2/87 = 1

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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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Algebra 2

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 \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: $\newline$ $(x - 0)^2/5^2 + (y - 0)^2/\sqrt{87}^2 = 1$ $\newline$ $x^2/25 + y^2/87 = 1$