How do you find the focus, directrix and sketch #x=-y^2+y+1#?

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Answer 1 · 2018-02-20T17:13:44

The Focus : #(3/2,1/2)#.

The eqn. of directrix : #x-1=0#.

graph{x=-y^2+y+1 [-11.25, 11.25, -5.62, 5.62]}

#:. y^2-y=-x+1#

# y^2-2*1/2*y+(1/2)^2=-x+1+(1/2)^2, i.e., #

# (y-1/2)^2=-x+5/4=-(x-5/4)=-4*1/4(x-5/4)#.

Shifting the origin to the point #(5/4,1/2)#, suppose that,

the new co-ords. of #(x,y)# become #(X,Y)#.

Then, #x=X+5/4, y=Y+1/2#.

So, the eqn. in #(X,Y)# system becomes, #Y^2=-4aX, a=1/4#.

This represents a parabola.

Now, recall the followings for the parabola #Y^2=-4aX : #

#"Focus "F(a,0)," and eqn. of directrix : "X+a=0#.

The focus #F" in "(X,Y)" is "X=a=1/4, Y=0#.

#:.F(x,y)" is "x-5/4=1/4, y-1/2=0; i.e., x=3/2,y=1/2#.

Thus, the focus #F(x,y)=F(3/2,1/2)#.

The eqn. of directrix is #(x-5/4)+1/4=0, i.e., x-1=0#.