How do you find the vertex, focus and directrix of # x=1/24y^2#?

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Answer 1 · 2018-01-18T12:56:05

Vertex is at #(0,0)#, focus is at #(6,0)# and directrix is #x=-6#

#x=1/24y^2 or y^2=4*6*x or(y-0)^2=4*6*(x-0) ; #

Comparing with equation for horizontal parabola #(a>0)#

openning right: #(y-k)^2=4a(x-h) ,(h,k)# being vertex , Here

#h=0 , k=0, a=6#. So vertex is at #(0,0)# . Focus for horizontal

parabola openning right: #y^2=4ax# is #(a,0)# . So focus

is at #(6,0)# Directrix for horizontal parabola opening right :

#y^2=4ax# is #x=-a :.#Directrix is #x=-6#

graph{x=y^2/24 [-80, 80, -40, 40]}