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 \\ (y^2-4y+2)/(4)+(x^2-8x+16)/(4*3)=0 \\ (\left(y-2\right)^2)/(2^2)-(\left(x-4\right)^2)/(\left(2√(3)\right)^2)=1 \\ \text{Compare it with }(\left(y-k\right)^2)/(a^2)-(\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)