Conics Question: Graph the equation x=-3y^2+12y+13 What are all applicable points? (vertex, focus, etc.)

1 Answer
Shwetank Mauria
Feb 12, 2017

Vertex is #(25,3)# and focus is #x=25 1/12#

Explanation:

When the equation of a parabola is of the form #x=a(y-k)^2+h#, the vertex is #(h,k)#, axis of symmetry is #y-k=0# and focus is #(h+1/(4a),k)# and directrix is #x=h-1/(4a)#

As #x=-3y^2+12y+13# is a quadratic equation and can be converted into vertex form as follows

#x=-3y^2+12y+13#

= #-3(y^2-4y+4-4)+12#

= #-3(y-2)^2+12+13#

= #-3(y-2)^2+25#

Its vertex is #(25,3)# and axis of symmetry is #y=2#

Its focus is #(25+1/(4xx(-3)),2)# i.e. #(24 11/12,2)# and

diectrix is #x=25-1/(4xx(-3))=25 1/12#
graph{x=-3y^2+12y+13 [-39.83, 40.17, -20.16, 19.84]}

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