How do you write the standard equation for a parabola with the given vertex (3,3) and focus: (-2,3)?

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Answer 1 · 2017-03-31T12:35:26

Mar 31, 2017

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Answer 2 · 2017-03-31T13:36:46

#x=-y^2/20+(3y)/10+51/20#

Here as the ordinate of vertex and focus is same , axis of symmetry is #y=3# and equation of parabola is of the form

#4p(x-3)=(y-3)^2#

As such its focus is #(3+p,3)# and as focus is #(-2,3)#

we have #p=-5# and equation of parabola is

#-20(x-3)=(y-3)^2#

or #x=-1/20(y^2-6y+9)+3#

or #x=-y^2/20+(3y)/10+51/20#

graph{x=-y^2/20+(3y)/10+51/20 [-52.5, 27.5, -16.64, 23.36]}