Question 11.3.4: Converting to Standard Form Find the center, vertices, and f...
Converting to Standard Form
Find the center, vertices, and foci of the ellipse with equation
3x² + 4y² + 12x – 8y – 32 = 0.
Complete squares on x and y.
\begin{aligned}3 x^{2}+4 y^{2}+12 x-8 y-32 &=0 & & \text { Original equation } \\\left(3 x^{2}+12 x\right)+\left(4 y^{2}-8 y\right) &=32 & & \text { Group like terms. } \\3\left(x^{2}+4 x\right)+4\left(y^{2}-2 y\right) &=32 & & \text { Factor out } 3 \text { and } 4 .\end{aligned}
3\left(x^{2}+4 x+4\right)+4\left(y^{2}-2 y+1\right)=32+12+4 Complete squares on x and y: add 3⋅4 and 4⋅1 to each side.
\begin{array}{ll}3(x+2)^{2}+4(y-1)^{2}=48 & \text { Factor and simplify. } \\\frac{(x+2)^{2}}{16}+\frac{(y-1)^{2}}{12}=1 & \text { Divide both sides by 48. (16 } \\ &\text { is the larger denominator. })\end{array}
The last equation is the standard form for a horizontal ellipse with center (-2, 1), a^{2}=16, b^{2}=12, \text { and } c^{2}=a^{2}-b^{2}=16-12=4 \text {. Thus, } a=4, b=\sqrt{12}=2 \sqrt{3}, and c = 2.
From the table, we have the following information:
The length of the major axis is 2a = 8.
The length of the minor axis is 2 b=4 \sqrt{3}.
Center: (h, k)=(-2,1)
Foci:
\begin{aligned}(h \pm c, k) &=(-2 \pm 2,1) \\&=(-4,1) \text { and }(0,1)\end{aligned}
Vertices:
\begin{aligned}(h \pm a, k) &=(-2 \pm 4,1) \\&=(-6,1) \text { and }(2,1)\end{aligned}
Endpoints of minor axis:
\begin{aligned}(h, k \pm b) &=(-2,1 \pm 2 \sqrt{3}) \\&=(-2,1+2 \sqrt{3}) \text { and }(-2,1-2 \sqrt{3}) \\& \approx(-2,4.46) \text { and }(-2,-2.46)\end{aligned}
The graph of the ellipse is shown in Figure 19.
