Write an equation for an ellipse centered at the origin, which has foci at (+-5,0) and vertices at (+-sqrt41,0).

Given Data: We are given the foci at (±5, 0) and vertices at (±√41, 0). Since the foci and vertices are on the x-axis, this indicates a horizontal orientation for the ellipse.Calculate 'a': The distance from the center to a vertex is represented by 'a', and the distance from the center to a focus is represented by 'c'. Using the given vertices, we can find the value of 'a'. a = √41Calculate 'c': Using the given foci, we can find the value of 'c'. c = 5Use Relationship Equation: The relationship between 'a', 'b', and 'c' for an ellipse is given by the equation c² = a² - b². We can use this to find the value of 'b'. c² = a² - b² 5² = (√41)² - b² 25 = 41 - b² b² = 41 - 25 b² = 16Find 'b': Now we can find the value of 'b' by taking the square root of b². b = √16 b = 4Write Standard Form: With the values of 'a' and 'b', we can write the standard form equation of the ellipse. Since the ellipse is horizontally oriented and centered at the origin, the standard form is: (x²/a²) + (y²/b²) = 1Substitute Values: Substitute the values of 'a' and 'b' into the equation. (x²/√41²) + (y²/4²) = 1 (x²/41) + (y²/16) = 1

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Write an equation for an ellipse centered at the origin, which has foci at
(+-5,0) and vertices at
(+-sqrt41,0).

Write an equation for an ellipse centered at the origin, which has foci at $( \pm 5,0)$ and vertices at $( \pm \sqrt{41}, 0)$ .

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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

( \pm 5,0)

and vertices at

( \pm \sqrt{41}, 0)

Given Data: We are given the foci at $(\pm 5, 0)$ and vertices at $(\pm\sqrt{41}, 0)$ . Since the foci and vertices are on the x-axis, this indicates a horizontal orientation for the ellipse.
Calculate $a$ : The distance from the center to a vertex is represented by $a$ , and the distance from the center to a focus is represented by $c$ . Using the given vertices, we can find the value of $a$ . $a = \sqrt{41}$
Calculate 'c': Using the given foci, we can find the value of 'c'. $\newline$ $c = 5$
Use Relationship Equation: The relationship between $a$ , $b$ , and $c$ for an ellipse is given by the equation $c^2 = a^2 - b^2$ . We can use this to find the value of $b$ .
$c^2 = a^2 - b^2$
$5^2 = (\sqrt{41})^2 - b^2$
$25 = 41 - b^2$
$b^2 = 41 - 25$
$b^2 = 16$
Find 'b': Now we can find the value of 'b' by taking the square root of $b^2$ . $b = \sqrt{16}$ $b = 4$
Write Standard Form: With the values of $a$ and $b$ , we can write the standard form equation of the ellipse. Since the ellipse is horizontally oriented and centered at the origin, the standard form is: $\newline$ $(\frac{x^2}{a^2}) + (\frac{y^2}{b^2}) = 1$
Substitute Values: Substitute the values of $a$ and $b$ into the equation. $(\frac{x^2}{\sqrt{41}^2}) + (\frac{y^2}{4^2}) = 1$ $(\frac{x^2}{41}) + (\frac{y^2}{16}) = 1$