How do you identify the vertex, focus, directrix and the length of the latus rectum and graph #3(y-3)=(x+6)^2#?

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Answer 1 · 2017-08-25T19:18:42

Given: #3(y-3)=(x+6)^2#

Divide both sides of the equation by 3:

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

Add 3 to both sides:

#y=1/3(x+6)^2+3#

Write the square in #(x - a)^2# form:

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

This equation is now in vertex form #y = 1/(4f)(x - h)^2+k#

where:

The vertex is

#(h,k) = (-6,3)#

The length, #l#, of the latus rectum is

#l = 4f#

#l = 3#

The focus is:

#(h, k+f) = (-6,15/4)#

The equation of the directrix is:

#y = k -f#

#y = 9/4#

Here is the graph:

graph{3(y-3)=(x+6)^2 [-16.83, 3.17, 1, 11]}