Ellipses: Learn It 4

Giselle Miller

Ellipses: Learn It 4

Graphing an Ellipse in General Form

Graphing an ellipse that’s not in its simplest form might seem challenging at first, but with a few steps, it becomes manageable and even fun! When you start with a more complex form of an ellipse, the key is to transform it into the standard form. This transformation process often involves completing the square and sometimes adjusting the coordinate system.

general form of an ellipse

[latex]a{x}^{2}+b{y}^{2}+cx+dy+e=0[/latex]

To rewrite the equation into standard form:

Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation
Factor out the coefficients of the [latex]x^2[/latex] and [latex]y^2[/latex] terms in preparation for completing the square.
Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant [latex]a(x-h)^2+b(y-k)^2 = \text{constant}[/latex].
Divide both sides of the equation by the constant term to express the equation in standard form.

To write the equation of an ellipse in standard form, you’ll need to complete the square two times.Given an expression of the form [latex]a\left(x^2+bx\right)[/latex], add [latex]\left(\dfrac{b}{2}\right)^2[/latex] inside the parentheses, then subtract [latex]a\left(\dfrac{b}{2}\right)^2[/latex] to counteract the change you made.If completing the square on one side of an equation, you may either subtract the value of [latex]a\left(\dfrac{b}{2}\right)^2[/latex] from that side, or add it to the other to maintain equality.

Then factor the perfect square trinomial you created inside the original parentheses.

Example

[latex]\qquad a\left(x^2+bx\right)[/latex]

[latex]=a\left(x^2+bx+ \left(\dfrac{b}{2}\right)^2\right)-a\left(\dfrac{b}{2}\right)^2[/latex]

[latex]=a\left(x+ \dfrac{b}{2}\right)^2-a\left(\dfrac{b}{2}\right)^2[/latex]

Graph the ellipse given by the equation [latex]4{x}^{2}+9{y}^{2}-40x+36y+100=0[/latex]. Identify and label the center, vertices, co-vertices, and foci.

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.