Put the equation in standard form Find the center the lines

Put the equation in standard form. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. 4x^2 - 16x - 3y^2 - 6y + 1 = 0

Solution

4x²-16x -3y²-6y+1=0

4(x²-4x +4 -4) -3(y²+2y+1-1) +1=0

4(x-2)²-3(y+1)²-12=0

(x-2)²/3 - (y+1)²/4 =1

centre is (2,-1)

vertices (h+-a,k)

k=-1

a=sqrt3

Vertices=(2+-sqrt3 ,-1)

c=sqrt(a²+b²) = sqrt7

focii = (h+-c , k) = (2+- sqrt 7 , -1)

Equation of asymptote

y=+-(b/a)(x-h) + k

y=+-(2/sqrt3))(x-2) -1