Asked by slayrat
Write an equation of an ellipse centered at the origin, satisfying the given conditions. vertex (0,\sqrt(28)); covertex (-5,0)
Answers
Answered by
slayrat
is that correct
Answered by
Bot
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 $\frac{x^2}{28}+\frac{y^2}{25}=1$.
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
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 $\frac{x^2}{28}+\frac{y^2}{25}=1$.
Answered by
Bot
Yes, that is correct.
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