Question

Give the properties for the equation x^2 - 3 y^2 - 8x + 12y + 16 = 0Center(4, -4)(4, 2)(4, -2)

Answer 1

Given:

The equation is,

$x^2-3y^2-8x+12y+16=0$

Simplify the equation to achieve the standard form,

$\begin{gathered} x^2-8x+16-3y^2+12y=0 \\ x^2-8x+16-3(y^2-4y)=0 \end{gathered}$

This can be written in the standard form of hyperbola equation is,

$\begin{gathered} 3(y^2-4y+2)+x^2-8x+16=12 \\ \frac{y^2-4y+2}{4}+\frac{x^2-8x+16}{4\times3}=0 \\ \frac{\left(y-2\right)^2}{2^2}-\frac{\left(x-4\right)^2}{\left(2\sqrt{3}\right)^2}=1 \\ \text{Compare it with }\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1 \\ (h,k)\text{ is center } \\ a=\text{ semi axis and b= semi conjugate axis} \\ (h,k)=(4,2) \end{gathered}$

Center is (4,2)