Question

Use Cartesian coordinates to express the equations for the ellipses determined by the conditions specified in Exercises 32–37.

$\text{foci}(3,\pm 5)\text{, major axis}12$

Short answer

The equation is$\frac{y^2}{36}+\frac{(x-3{)}^2}{11}=1.$

Step-by-step solution

Step 1. Given information.

It is given,

$\text{foci}(3,\pm 5)\text{, major axis}12$

Step 2. Value of the variables.

Focus points are $(3,5)(3,-5).$

$\text{center}=\left(\frac{3+3}{2},\frac{5-5}{2}\right)=(3,0)$

$$\begin{gathered} Givenmajoraxisis12 \\ 2b=12 \\ b=6 \\ c=\sqrt{(3-3{)}^2+(0-5{)}^2}\left[\text{since}D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right] \\ \text{That is the distance from}(3,5)(3,0)\text{.} \\ c=5 \\ Now, \\ {c}^2=b^2-a^2 \\ {5}^2=6^2-a^2 \\ 25=36-a^2 \\ {a}^2=11 \end{gathered}$$

Step 3. Substitution.

On substituting the obtained values in,

$$\begin{gathered} \frac{(y-k{)}^2}{b^2}+\frac{(x-h{)}^2}{a^2}=1\text{where}b>a,(h,k)\text{is the center.} \\ \frac{(y-0{)}^2}{36}+\frac{(x-3{)}^2}{11}=1 \\ \frac{y^2}{36}+\frac{(x-3{)}^2}{11}=1 \end{gathered}$$