Question 828448
{{{x=ay^2+h}}}
(The value for "k" is 0).


Vertex is ON the origin, so then h=0 also.  Vertex would still be some (h,k), from a fuller standard form, {{{x=a(y-k)^2+k}}}.  


What is the value of a?  
Point (2,3) gives {{{3=a(2)^2}}}
{{{3=a*4}}}
{{{a=3/4}}}


Your equation, {{{highlight(x=(3/4)y^2)}}}