What is the vertex form of #x = (12y - 3)^2 -144x+1#?

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Answer 1 · 2018-02-16T12:24:14

The vertex is at #(1/145,1/4)# and vertex form of equation
is #x=144/145(y-1/4)^2+1/145#

#x=(12y-3)^2-144x+1 or 145x=(12y-3)^2+1# or

#145x=144(y-1/4)^2+1 or x=144/145(y-1/4)^2+1/145#

The vertex form of equation is #x = a (y - k)^2 + h#

If a is positive the parabola opens right , if a is negative the

parabola opens left . Vertex: #(h, k); h=1/145, k =1/4,a=144/145#

The vertex is at #(1/145,1/4)# and vertex form of equation

is #x=144/145(y-1/4)^2+1/145#

graph{x=144/145(y-1/4)^2+1/145 [-10, 10, -5, 5]} [Ans]