Find center, foci, and vertices of ellipse (x+3)^2/21+(y-5)^2/25=1

I don’t know if we can find the foci of this ellipse, but we can find the centre and the vertices. First of all, let us state the standard equation of an ellipse.

(If there is a way to solve for the foci of this ellipse, please let me know! I am learning this stuff currently. )

$$\frac{(x-x_{1})^2}{a^2}+ \frac{(y-y_{1})^2}{b^2}=1$$

Where $$(x_{1}, y_{1})$$ is the centre of the ellipse. Just by looking at your equation right away, we can tell that the centre of the ellipse is:

$$(-3,5)$$

Now to find the vertices, we must first remember that the vertices of an ellipse are on the major axis.

The major axis in this case is that of the y-axis. In other words,

$$b^2> a^2$$

So we know that b=5 from your equation given. The vertices are 5 away from the centre, so we find that the vertices of your ellipse are:

$$(-3,10)$$
& $$(-3,0)$$

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