Question

complete the square to find and write the standard form for the hyperbola (y^{2}+8y - 3x^{2}+12x + 13 = 0)

Explanation:

Step1: Group x and y terms

$(y^{2}+8y)-(3x^{2}-12x)+13 = 0$

Step2: Factor out coefficients for x - terms

$(y^{2}+8y)-3(x^{2}-4x)+13 = 0$

Step3: Complete the square for y - terms

For $y^{2}+8y$, add and subtract $(\frac{8}{2})^2 = 16$. So we have $(y^{2}+8y + 16-16)-3(x^{2}-4x)+13 = 0$, which is $(y + 4)^{2}-16-3(x^{2}-4x)+13 = 0$.

Step4: Complete the square for x - terms

For $x^{2}-4x$, add and subtract $(\frac{-4}{2})^2=4$. So we get $(y + 4)^{2}-16-3(x^{2}-4x + 4-4)+13 = 0$, which is $(y + 4)^{2}-16-3((x - 2)^{2}-4)+13 = 0$.

Step5: Expand and simplify

$(y + 4)^{2}-16-3(x - 2)^{2}+12 + 13 = 0$.
$(y + 4)^{2}-3(x - 2)^{2}+9 = 0$.
Move 9 to the other side: $3(x - 2)^{2}-(y + 4)^{2}=9$.

Step6: Divide by 9 to get standard form

$\frac{(x - 2)^{2}}{3}-\frac{(y + 4)^{2}}{9}=1$

Answer:

$\frac{(x - 2)^{2}}{3}-\frac{(y + 4)^{2}}{9}=1$