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]\dfrac{{x}^{2}}{{b}^{2}}+\dfrac{{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{align}c&=\pm \sqrt{{a}^{2}-{b}^{2}} \\ &=\pm \sqrt{25 - 9} \\ &=\pm \sqrt{16} \\ &=\pm 4 \end{align}[/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{gathered} 4{x}^{2}+25{y}^{2}=100 \\[1.5mm] \dfrac{4{x}^{2}}{100}+\dfrac{25{y}^{2}}{100}=\dfrac{100}{100} \\[1.5mm] \dfrac{{x}^{2}}{25}+\dfrac{{y}^{2}}{4}=1 \end{gathered}[/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]\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{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{align}c&=\pm \sqrt{{a}^{2}-{b}^{2}} \\ &=\pm \sqrt{25 - 4} \\ &=\pm \sqrt{21} \end{align}[/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]\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}+\dfrac{{\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{align}c&=\pm \sqrt{{a}^{2}-{b}^{2}} \\ &=\pm \sqrt{9 - 4} \\ &=\pm \sqrt{5} \end{align}[/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]\dfrac{{\left(x - 5\right)}^{2}}{9}+\dfrac{{\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]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}+\dfrac{{\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{align}c&=\pm \sqrt{{a}^{2}-{b}^{2}} \\ &=\pm \sqrt{9 - 4} \\ &=\pm \sqrt{5} \end{align}[/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.