Question

I need this practice problem answered I will provide another picture with the answer options My subject right now is pre calculus

Answer 1

The equation of an ellipse is given as;

$\frac{(x-6)^2}{36}+\frac{(y+3)}{100}=1$

But, the equation of an ellipse with centre (h,k), a is the distance from the center to the end of the major axis and b is the distance from the center to the end of the minor axis is;

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

Now, we relate to the given equation to get the ventre (h,k), we have;

$\begin{gathered} x-h=x-6 \\ h=6 \\ \text{and} \\ y-k=y+3 \\ k=-3 \\ (h,k)=(6,-3) \end{gathered}$

Also, the distance from the center to the end of the major axis is;

$\begin{gathered} a^2=100 \\ a=10\text{units} \end{gathered}$

The endpoints of the major axis are 10 units from the center.

The distance from the center to the end of the minor axis is;

$\begin{gathered} b^2=36 \\ b=6\text{units} \end{gathered}$

The endpoints of the minor axis are 6 units from the center.