Answer
$\frac{(x-3)^2}{25}+\frac{(y+5)^2}{4}=1$
Ellipse: center $C(3, -5)$, foci $F(3\pm\sqrt {21}, -5)$, vertices $V_1(-2,-5)$ and $V_2(8,-5)$
lengths of the major axis $10$, length of the minor axis $4$
See graph.
Work Step by Step
Step 1. Form squares with the variables: $4(x^2-6x+9)+25(y^2+10y+25)=36+625-561$ or $4(x-3)^2+25(y+5)^2=100$. Divide both sides by 100, we get $\frac{(x-3)^2}{25}+\frac{(y+5)^2}{4}=1$
Step 2. We can identify the above equation as an ellipse with a center at $C(3, -5)$
Step 3. With $a=5, b=2$, we have $c=\sqrt {25-4}=\sqrt {21}$
Step 4. We can find the foci at $F(3\pm\sqrt {21}, -5)$
Step 5. The vertices are $(3\pm5, -5)$ or $V_1(-2,-5)$ and $V_2(8,-5)$
Step 6. The lengths of the major axis is $2a=10$ and the length of the minor axis is $2b=4$
Step 7. See graph.
