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

##### 1 Answer

Vertex

Focus

Directrix

#### Explanation:

Given -

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

We have to find the vertex, focus and directrix.

The given equation is in the form.

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

It is in the vertex form. Then its vertex is

Obtain the values of

#h= 1# x-coordinate of the vertex

#k=-1# y coordinate of the vertex

Then the vertex of the given equation is

If the term

In the given equation

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

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

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

One is equal to

So we can replace one with

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

Now we have the constant term 4.

The distance between vertex and focus or vertex and directrix is

We can find the focus and directrix.

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

Focus is

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