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

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Answer 1 · 2017-10-01T10:42:08

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

Directrix would be #y= 21/20#

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#