for the parabola, x^2+6 x-12 y-51 = 0 find the vertex,focus and the directrix

Question

find the vertex,the focus and the directrix of the parabola x^2+6 x-12 y-51 = 0

Answer 1

Given equation x ² + 6x  - 12y - 51 = 0

We first put the equation in to the form for a translated parabola (x - h )² = 4p (y - k)

To do this we complete the squre on the x  terms and move the other terms to right.

x ² + 6x = 12y + 51

To change the expression into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

x coefficient is 6,(half the x coefficient)² = 9

x ² + 6x + 9 = 12y + 51 + 9

(x + 3)² = 12y + 60

In the next step we factored 12 from the right hand side to make the coefficient of our y  "+1", as this is the standard form.

(x + 3)² = 12(y + 5)

(x -( -3))² = 4(3)(y - (-5))

Compare it to  parabola equation is (x - h)² = 4p (y - k), where (h, k) = vertex and p = directed distance from vertex to focus.

p = 3

Vertex of parabola  = (-3, -5)

Focus = (h , k + p ) = [-3, -5+3)] = (-3, -2)

Equation of directrix is y  = k - p

y  = -5 -3

y  = -8

Vertex (-3, -5), focus (-3, -2) and directrix  y = -8.