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Sagot :
Step 1
Given;
[tex]4x^2-y^2+24x+4y+28=0[/tex]Required; To find the center that eliminates the linear terms
Step 2
[tex]\begin{gathered} 4x^2-y^2+24x+4y=-28 \\ 4x^2+24x-y^2+4y=-28 \\ Complete\text{ the square }; \\ 4x^2+24x \\ \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=4 \\ b=24 \\ c=0 \end{gathered}[/tex][tex]\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\frac{24}{2\times4} \\ d=\frac{24}{8} \\ d=3 \end{gathered}[/tex][tex]\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{24^2}{4\times4} \\ e=0-\frac{576}{16}=-36 \end{gathered}[/tex]Step 3
Substitute a,d,e into the vertex form
[tex]\begin{gathered} a(x+d)^2+e \\ 4(x+_{}3)^2-36 \end{gathered}[/tex][tex]\begin{gathered} 4(x+3)^2-36-y^2+4y=-28 \\ 4(x+3)^2-y^2+4y=\text{ -28+36} \\ \\ \end{gathered}[/tex]Step 4
Completing the square for -y²+4y
[tex]\begin{gathered} \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=-1 \\ b=4 \\ c=0 \end{gathered}[/tex][tex]\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\text{ }\frac{4}{2\times-1} \\ d=\frac{4}{-2} \\ d=-2 \end{gathered}[/tex][tex]\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{4^2}{4\times(-1)} \\ \\ e=0-\frac{16}{-4} \\ e=4 \end{gathered}[/tex]Step 5
Substitute a,d,e into the vertex form
[tex]\begin{gathered} a(y+d)^2+e \\ =-1(y+(-2))^2+4 \\ =-(y-2)^2+4 \end{gathered}[/tex]Step 6
[tex]\begin{gathered} 4(x+3)^2-y^2+4y=\text{ -28+36} \\ 4(x+3)^2-(y-2)^2+4=-28+36 \\ 4(x+3)^2-(y-2)^2=-28+36-4 \\ 4(x+3)^2-(y-2)^2=4 \\ \frac{4(x+3)^2}{4}-\frac{(y-2)^2}{4}=\frac{4}{4} \\ (x+3)^2-\frac{(y-2)^2}{2^2}=1 \end{gathered}[/tex]Step 7
[tex]\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \text{This is the }form\text{ of a hyperbola.} \\ \text{From here } \\ a=1 \\ b=2 \\ k=2 \\ h=-3 \end{gathered}[/tex]Hence the answer is (-3,2)
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