Finding the equation of an ellipse

Question

Given : a vertex $(18,-2)$, a focus $(12,-2)$ and a minor axis length of $24$ How do you solve for this one ?

Answer 1

$$(\frac{x-3}{15})^2+(\frac{y+2}{12})^2=1$$ fits the bill.

The template $$(\frac{x-c_1}{a})^2+(\frac{y-c_2}{b})^2=1$$ works since the major axis is parallel to the $x$-axis; defined as it is by the vertex and focus.

$b=12$ since the minor axis is $24$ long.

The distance from the center to the focus is $c=a-6$.

$c^2=a^2-b^2$ then gives $(a-6)^2=a^2-12^2$ i.e. $a=15$.

The vertex and focus line goes through the center, making the y coordinate of the center $-2$, and the equation becomes

$$(\frac{x-c_1}{15})^2+(\frac{y+2}{12})^2=1$$

Substituting $(18,-2)$ which is a point on the ellipse we get $(\frac{18-c_1}{15})^2+(\frac{-2+2}{12})^2=1$ which gives $|18-c_1|=15$ making $c_1=3$ or $33$ but $33$ yields an ellipse for which $(12,-2)$ is not a focus.

Answer 2

Hint:

from the figure: $a=\overline{CA}$, $b=\overline{CB}$, $c=\overline {CF}$. For an ellipse we have: $a^2-b^2=c^2$ that, for a minor axis length of $24$ becomes $a^2-12^2=c^2$ and in your case we know also $a=6+c$. Substitute $a$ and solve $$ (6+c)^2-12^2=c^2 $$