Question

find the center of this hyperbola.

-9x² + 4y² - 36x - 16y - 164 = 0

center : (, )

Explanation:

Step1: Group x and y terms

Group \(x\) - terms \(-9x^{2}-36x\) and \(y\) - terms \(4y^{2}-16y\) together:
\(-9(x^{2} + 4x)+4(y^{2}-4y)-164 = 0\)

Step2: Complete the square for x

For \(x^{2}+4x\), using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\), \(2b = 4\Rightarrow b = 2\), \(x^{2}+4x=(x + 2)^{2}-4\)
For \(y^{2}-4y\), using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=y\), \(2b = 4\Rightarrow b = 2\), \(y^{2}-4y=(y - 2)^{2}-4\)
The equation becomes:
\(-9((x + 2)^{2}-4)+4((y - 2)^{2}-4)-164 = 0\)

Step3: Expand and simplify

\(-9(x + 2)^{2}+36+4(y - 2)^{2}-16-164 = 0\)
\(-9(x + 2)^{2}+4(y - 2)^{2}-144 = 0\)
\(4(y - 2)^{2}-9(x + 2)^{2}=144\)
\(\frac{(y - 2)^{2}}{36}-\frac{(x + 2)^{2}}{16}=1\)

Answer:

\((-2,2)\)