What are the vertex, focus and directrix of # y=(x-1)^2-1 #?

1 Answer
Nov 9, 2017

Vertex #(1,-1)#

Focus #(1, -0.75)#
Directrix #y=-1.25#

Explanation:

Given -

#y=(x-1)^2-1#

We have to find the vertex, focus and directrix.

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The given equation is in the form.

#y=a(x-h)^2+k#

It is in the vertex form. Then its vertex is #(h, k)#

Obtain the values of #h# and #k# from the given equation.

#h= 1# x-coordinate of the vertex
#k=-1# y coordinate of the vertex

Then the vertex of the given equation is #(1, -1)#

If the term #a# in the equation #y=a(x-h)^2+k# is positive, it means the curve opens up.

In the given equation #a=1#. one is a positive value, hence the parabola of the given equation opens up.

The standard form of a parabola of this type is as follows.
[for a parabola whose vertex is away from origin]

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

Where

4 is a constant

#a# is the distance between vertex and focus.

We shall rewrite the given equation like this

#y=(x-1)^2-1#

#(x-1)^2-1=y#

#(x-1)^2=y+1#

It can be written as

#(x-1)^2=(y+1)#

It appears, there is no #4a# term.We shall bring it like this.

#(x-1)^2=(y+1)# this equation can be written like this

#(x-1)^2=1 xx (y+1)#

One is equal to #4 xx 1/4#

So we can replace one with #4 xx 1/4# and rewrite

#(x-1)^2=4 xx1/4xx (y+1)#

Now we have the constant term 4. #1/4 # is #a#

The distance between vertex and focus or vertex and directrix is #a=1/4#

We can find the focus and directrix.

The point which lies vertically at a distance of 0.25 above vertex is focus.

Focus is #(1, (-1+0.25))#; #(1, -0.75)#

Find the point the lies vertically at a distance of 0.25 below vertex.
Use its y- coordinate to find the equation of directrix.

#(1, (-1 + (-25))) #
#(1, (-1 -25)) #
#(1, -1.25) #

Directrix #y=-1.25#