The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form

[latex]\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1[/latex]

First, we identify the center, [latex]\left(h,k\right)[/latex]. The center is halfway between the vertices, [latex]\left(-2,-8\right)[/latex] and [latex]\left(-2,\text{2}\right)[/latex]. Applying the midpoint formula, we have:

[latex]\begin{array}{l}\left(h,k\right)=\left(\frac{-2+\left(-2\right)}{2},\frac{-8+2}{2}\right)\hfill \\ \text{ }=\left(-2,-3\right)\hfill \end{array}[/latex]

Next, we find [latex]{a}^{2}[/latex]. The length of the major axis, [latex]2a[/latex], is bounded by the vertices. We solve for [latex]a[/latex] by finding the distance between the y-coordinates of the vertices.

[latex]\begin{array}{c}2a=2-\left(-8\right)\\ 2a=10\\ a=5\end{array}[/latex]

So [latex]{a}^{2}=25[/latex].

Now we find [latex]{c}^{2}[/latex]. The foci are given by [latex]\left(h,k\pm c\right)[/latex]. So, [latex]\left(h,k-c\right)=\left(-2,-7\right)[/latex] and [latex]\left(h,k+c\right)=\left(-2,\text{1}\right)[/latex]. We substitute [latex]k=-3[/latex] using either of these points to solve for [latex]c[/latex].

[latex]\begin{array}{c}k+c=1\\ -3+c=1\\ c=4\end{array}[/latex]
So [latex]{c}^{2}=16[/latex].

Next, we solve for [latex]{b}^{2}[/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex].

[latex]\begin{array}{c}{c}^{2}={a}^{2}-{b}^{2}\\ 16=25-{b}^{2}\\ {b}^{2}=9\end{array}[/latex]

Finally, we substitute the values found for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form equation for an ellipse:

[latex]\frac{{\left(x+2\right)}^{2}}{9}+\frac{{\left(y+3\right)}^{2}}{25}=1[/latex]

a. We are assuming a horizontal ellipse with center [latex]\left(0,0\right)[/latex], so we need to find an equation of the form [latex]\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1[/latex], where [latex]a>b[/latex]. We know that the length of the major axis, [latex]2a[/latex], is longer than the length of the minor axis, [latex]2b[/latex]. So the length of the room, 96, is represented by the major axis, and the width of the room, 46, is represented by the minor axis.

• Solving for [latex]a[/latex], we have [latex]2a=96[/latex], so [latex]a=48[/latex], and [latex]{a}^{2}=2304[/latex].
• Solving for [latex]b[/latex], we have [latex]2b=46[/latex], so [latex]b=23[/latex], and [latex]{b}^{2}=529[/latex].

Therefore, the equation of the ellipse is [latex]\frac{{x}^{2}}{2304}+\frac{{y}^{2}}{529}=1[/latex].

b. To find the distance between the senators, we must find the distance between the foci, [latex]\left(\pm c,0\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex]. Solving for [latex]c[/latex], we have:

[latex]\begin{array}{ll}{c}^{2}={a}^{2}-{b}^{2}\hfill & \hfill \\ {c}^{2}=2304 - 529\hfill & \begin{array}{cccc}& & & \end{array}\text{Substitute using the values found in part (a)}.\hfill \\ c=\pm \sqrt{2304 - 529}\hfill & \begin{array}{cccc}& & & \end{array}\text{Take the square root of both sides}.\hfill \\ c=\pm \sqrt{1775} \hfill & \begin{array}{cccc}& & & \end{array}\text{Subtract}.\hfill \\ c\approx \pm 42\hfill & \begin{array}{cccc}& & & \end{array}\text{Round to the nearest foot}.\hfill \end{array}[/latex]

The points [latex]\left(\pm 42,0\right)[/latex] represent the foci. Thus, the distance between the senators is [latex]2\left(42\right)=84[/latex] feet.

First, we determine the position of the major axis. Because [latex]25>9[/latex], the major axis is on the y-axis. Therefore, the equation is in the form [latex]\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1[/latex], where [latex]{b}^{2}=9[/latex] and [latex]{a}^{2}=25[/latex]. It follows that:

• the center of the ellipse is [latex]\left(0,0\right)[/latex]
• the coordinates of the vertices are [latex]\left(0,\pm a\right)=\left(0,\pm \sqrt{25}\right)=\left(0,\pm 5\right)[/latex]
• the coordinates of the co-vertices are [latex]\left(\pm b,0\right)=\left(\pm \sqrt{9},0\right)=\left(\pm 3,0\right)[/latex]
• the coordinates of the foci are [latex]\left(0,\pm c\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex] Solving for [latex]c[/latex], we have:

[latex]\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}}\hfill \\ =\pm \sqrt{25 - 9}\hfill \\ =\pm \sqrt{16}\hfill \\ =\pm 4\hfill \end{array}[/latex]

Therefore, the coordinates of the foci are [latex]\left(0,\pm 4\right)[/latex].

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

First, use algebra to rewrite the equation in standard form.

[latex]\begin{array}{l} 4{x}^{2}+25{y}^{2}=100\hfill \\ \text{ }\frac{4{x}^{2}}{100}+\frac{25{y}^{2}}{100}=\frac{100}{100}\hfill \\ \text{ }\frac{{x}^{2}}{25}+\frac{{y}^{2}}{4}=1\hfill \end{array}[/latex]

Next, we determine the position of the major axis. Because [latex]25>4[/latex], the major axis is on the x-axis. Therefore, the equation is in the form [latex]\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1[/latex], where [latex]{a}^{2}=25[/latex] and [latex]{b}^{2}=4[/latex]. It follows that:

• the center of the ellipse is [latex]\left(0,0\right)[/latex]
• the coordinates of the vertices are [latex]\left(\pm a,0\right)=\left(\pm \sqrt{25},0\right)=\left(\pm 5,0\right)[/latex]
• the coordinates of the co-vertices are [latex]\left(0,\pm b\right)=\left(0,\pm \sqrt{4}\right)=\left(0,\pm 2\right)[/latex]
• the coordinates of the foci are [latex]\left(\pm c,0\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex]. Solving for [latex]c[/latex], we have:

[latex]\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}}\hfill \\ =\pm \sqrt{25 - 4}\hfill \\ =\pm \sqrt{21}\hfill \end{array}[/latex]

Therefore the coordinates of the foci are [latex]\left(\pm \sqrt{21},0\right)[/latex].

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

First, we determine the position of the major axis. Because [latex]9>4[/latex], the major axis is parallel to the y-axis. Therefore, the equation is in the form [latex]\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1[/latex], where [latex]{b}^{2}=4[/latex] and [latex]{a}^{2}=9[/latex]. It follows that:

• the center of the ellipse is [latex]\left(h,k\right)=\left(-2,\text{5}\right)[/latex]
• the coordinates of the vertices are [latex]\left(h,k\pm a\right)=\left(-2,5\pm \sqrt{9}\right)=\left(-2,5\pm 3\right)[/latex], or [latex]\left(-2,\text{2}\right)[/latex] and [latex]\left(-2,\text{8}\right)[/latex]
• the coordinates of the co-vertices are [latex]\left(h\pm b,k\right)=\left(-2\pm \sqrt{4},5\right)=\left(-2\pm 2,5\right)[/latex], or [latex]\left(-4,5\right)[/latex] and [latex]\left(0,\text{5}\right)[/latex]
• the coordinates of the foci are [latex]\left(h,k\pm c\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex]. Solving for [latex]c[/latex], we have:

[latex]\begin{array}{l}\begin{array}{l}\\ c=\pm \sqrt{{a}^{2}-{b}^{2}}\end{array}\hfill \\ =\pm \sqrt{9 - 4}\hfill \\ =\pm \sqrt{5}\hfill \end{array}[/latex]

Therefore, the coordinates of the foci are [latex]\left(-2,\text{5}-\sqrt{5}\right)[/latex] and [latex]\left(-2,\text{5+}\sqrt{5}\right)[/latex].

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

We must begin by rewriting the equation in standard form.
[latex]4{x}^{2}+9{y}^{2}-40x+36y+100=0[/latex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation.
[latex]\left(4{x}^{2}-40x\right)+\left(9{y}^{2}+36y\right)=-100[/latex]
Factor out the coefficients of the squared terms.
[latex]4\left({x}^{2}-10x\right)+9\left({y}^{2}+4y\right)=-100[/latex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[latex]4\left({x}^{2}-10x+25\right)+9\left({y}^{2}+4y+4\right)=-100+100+36[/latex]
Rewrite as perfect squares.
[latex]4{\left(x - 5\right)}^{2}+9{\left(y+2\right)}^{2}=36[/latex]
Divide both sides by the constant term to place the equation in standard form.
[latex]\frac{{\left(x - 5\right)}^{2}}{9}+\frac{{\left(y+2\right)}^{2}}{4}=1[/latex]
Now that the equation is in standard form, we can determine the position of the major axis. Because [latex]9>4[/latex], the major axis is parallel to the x-axis. Therefore, the equation is in the form [latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex], where [latex]{a}^{2}=9[/latex] and [latex]{b}^{2}=4[/latex]. It follows that:

• the center of the ellipse is [latex]\left(h,k\right)=\left(5,-2\right)[/latex]
• the coordinates of the vertices are [latex]\left(h\pm a,k\right)=\left(5\pm \sqrt{9},-2\right)=\left(5\pm 3,-2\right)[/latex], or [latex]\left(2,-2\right)[/latex] and [latex]\left(8,-2\right)[/latex]
• the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)=\left(\text{5},-2\pm \sqrt{4}\right)=\left(\text{5},-2\pm 2\right)[/latex], or [latex]\left(5,-4\right)[/latex] and [latex]\left(5,\text{0}\right)[/latex]
• the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex]. Solving for [latex]c[/latex], we have:

[latex]\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}}\hfill \\ =\pm \sqrt{9 - 4}\hfill \\ =\pm \sqrt{5}\hfill \end{array}[/latex]
Therefore, the coordinates of the foci are [latex]\left(\text{5}-\sqrt{5},-2\right)[/latex] and [latex]\left(\text{5+}\sqrt{5},-2\right)[/latex].

Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.