How do you find the vertex, focus and directrix of # y^2 = 4(x+3y)#?

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Answer 1 · 2016-09-12T10:37:24

Vertex : #(-9, 6)#
Focus : #(-8, 6)#
Directrix: #x = -10#

Given -

#y^2=4(x+3y)#

To find, Vertex, Focus and Directrix, we have to write the above equation in the form of -

#(y-k)^2=4p(x-h)#

#y^2=4x+12y#
#y^2-12y=4x#

Let us make the left side a perfect square. For this,we shall square half the coefficient of #y# and add it on both sides of the equation.

#y^2-12y+36=4x+36#

#(y-6)^2=4(x+9)#

And remember the equation resembles

#y^2=4px#

In our case #p=1# [Since 4 can be written as #4 xx 1#

Use the above information to get the required parameters.

Vertex : #(-9, 6)#
Focus : #(-8, 6)#
Directrix: #x = -10#

Look at the graph also