How do you write the parabola 25x^2+20x+5y-1=0 in standard form and find the vertex, focus, and directrix?

1 Answer
Oct 1, 2017

Vertex (-2/5,1)
Focus would be (-2/5, 19/20)

Directrix would be y= 21/20

Explanation:

First rewrite it as 5y= -25x^2 -20x+1
Or, y= -5x^2-4x+1/5
Now make a perfect square of x- terms as shown the steps below:

y= -5(x^2+4/5 x)+ +1/5

y= -5(x^2 +4/5 x +4/25 -4/25) + 1/5

y= -5(x^2 +4/5 x +4/25) + 4/5 +1/5

y= -5(x+2/5)^2 +1 . Rewrite further as (x +2/5)^2 = -1/5 y +1/5 or as (x+2/5)^2 = 4 (-1/20) (y-1)

This is the equation of the parabola in standard form. Its vertex is (-2/5, 1) Line of symmetry is x= -2/5

Focus would be (-2/5, 1-1/20) or (-2/5, 19/20)

Directrix would be y= 1 +1/20= 21/20