Write an equation of an ellipse centered at the origin, satisfying the given conditions. vertex (0,\sqrt(28)); covertex (-5,0)

The standard form equation of an ellipse centered at the origin is:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
where $a$ is the distance from the center to the vertex, and $b$ is the distance from the center to the co-vertex.

Using the given information, we know that:

$a=\sqrt{28}$ (distance from center to vertex)
$b=5$ (distance from center to co-vertex)

Substituting these values into the standard form equation, we get:

$\frac{x^2}{(\sqrt{28})^2}+\frac{y^2}{5^2}=1$

Simplifying, we can write the equation as:

$\frac{x^2}{28}+\frac{y^2}{25}=1$

Therefore, the equation of the ellipse is $\frac{x^2}{28}+\frac{y^2}{25}=1$.

is that correct

Yes, that is correct.

To find the equation of an ellipse centered at the origin with the given conditions, we need to determine the values of a and b.

Given that the vertex is at (0, √28) and the covertex is at (-5, 0), we can determine the values of a and b as follows:

The distance from the origin to the vertex is equal to a, and the distance from the origin to the covertex is equal to b.

The distance from the origin to the vertex is √(0^2 + (√28)^2) = √(0 + 28) = √28 = 2√7.

So, a = 2√7.

The distance from the origin to the covertex is √((-5)^2 + 0^2) = √(25 + 0) = √25 = 5.

So, b = 5.

Now, we can write the equation of the ellipse:

(x^2/a^2) + (y^2/b^2) = 1

Substituting the values of a and b:

(x^2/(2√7)^2) + (y^2/5^2) = 1

Simplifying:

(x^2/28) + (y^2/25) = 1

Therefore, the equation of the ellipse centered at the origin, satisfying the given conditions, is:

(x^2/28) + (y^2/25) = 1.