Given the equation of an ellipse: #41
[tex]x^2+8x+4y^2-40y+112=0[/tex]
To find the center and the vertices, we will complete the square of (x) and (y):
[tex]\begin{gathered} x^2+8x+4(y^2-10y)=-112 \\ (x^2+8x+16)+4(y^2-10y+25)=-112+16+100 \end{gathered}[/tex]
Factor for (x) and (y) then simplify:
[tex]\begin{gathered} (x+4)^2+4(y-5)^2=4\rightarrow(\div4) \\ \\ \frac{(x+4)^2}{4}+\frac{(y-5)^2}{1}=1 \end{gathered}[/tex]
The general form of the ellipse will be:
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]
Comparing the equation with the form:
[tex]\begin{gathered} (h,k)=(-4,5) \\ a=\sqrt[]{4}=2 \\ b=\sqrt[]{1}=1 \\ c=\pm\sqrt[]{a^2-b^2}=\pm\sqrt[]{4-1}=\pm\sqrt[]{3} \end{gathered}[/tex]
The graph of the ellipse will be as shown in the following picture:
As shown in the figure:
The center C = (h,k) = (-4 , 5)
The vertices denoted by V = (-6, 5) and (-2, 5)
Foci denoted by F =
[tex](-4+\sqrt[]{3},5),(-4-\sqrt[]{3},5)[/tex]
