Find the vertex, focus, and directrix of the conic section 2x^2+12x+16y+2=0

Question

i need to graph it as well.

Answer 1

Given equatioon 2x^2+12x+16y+2 = 0

Divide to each side by 2.

x^2+6x+8y+1 = 0

Add 8 to each side.

x^2+6x+8y+1+8 = 0+8

x^2+6x+9+8y = 8

Subtract 8x from each side.

x^2+6x+9 = -8y+8

(x+3)^2 = -8(y-1)

(x-(-3))^2 = 4(-2)(y-1)

Compare it standard form of parabola (x-h)^2 = 4a(y-k)

Vertex = (h,k) = (-3,1)

Axis of symmetry = x-h = 0

x-(-3) = 0

x = -3

Focus = (h,k+a)= (-3,1+(-2)) = (-3,-1)

Directrix = y-k+a = 0

y-1+(-2) = 0

y-3 = 0

y = 3