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Write an equation for an ellipse centered at the origin, which has foci at
(+-sqrt12,0) and vertices at
(+-sqrt37,0).

Write an equation for an ellipse centered at the origin, which has foci at $( \pm \sqrt{12}, 0)$ and vertices at $( \pm \sqrt{37}, 0)$ .

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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

( \pm \sqrt{12}, 0)

and vertices at

( \pm \sqrt{37}, 0)

.

Given Information: We are given: $\newline$ Foci: $(\pm\sqrt{12}, 0)$ $\newline$ Vertices: $(\pm\sqrt{37}, 0)$ $\newline$ Since the vertices and foci are on the x-axis, this indicates a horizontal orientation for the ellipse. $\newline$ The distance from the center to a vertex is $'a'$ , and the distance from the center to a focus is $'c'$ . $\newline$ We can determine the values of $'a'$ and $'c'$ from the given information. $\newline$ $a = \sqrt{37}$ $\newline$ $c = \sqrt{12}$
Find 'b': Next, we need to find the value of 'b', which is the distance from the center to the co-vertices. The relationship between ' $a$ ', ' $b$ ', and ' $c$ ' for an ellipse is given by the equation $c^2 = a^2 - b^2$ . We can rearrange this to solve for ' $b$ ': $b^2 = a^2 - c^2$
Calculate 'b': Now we will plug in the values of 'a' and 'c' to find 'b': $\newline$ $b^2 = (\sqrt{37})^2 - (\sqrt{12})^2$ $\newline$ $b^2 = 37 - 12$ $\newline$ $b^2 = 25$ $\newline$ Taking the square root of both sides, we get: $\newline$ $b = \sqrt{25}$ $\newline$ $b = 5$
Standard Form Equation: With the values of $a$ and $b$ , we can now write the equation of the ellipse in standard form. For a horizontal ellipse centered at the origin, the standard form is: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
Substitute Values: Substituting the values of $a$ and $b$ into the standard form equation, we get: $\left(\frac{x^2}{\sqrt{37}^2}\right) + \left(\frac{y^2}{5^2}\right) = 1$ Simplifying the equation, we have: $\left(\frac{x^2}{37}\right) + \left(\frac{y^2}{25}\right) = 1$

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