Question 1166292
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One problem per post!!!<br>
Here is all you need to work these problems yourself....<br>
Vertex form of equation of a parabola that opens up or down:<br>
{{{y-k=(1/(4p))(x-h)^2}}}
or
{{{y = (1/(4p))(x-h)^2+k}}}<br>
Vertex form of equation of a parabola that opens right or left:<br>
{{{x-h=(1/(4p))(y-k)^2}}}
or
{{{x = (1/(4p))(y-k)^2+h}}}<br>
In both formulas, the vertex is (h,k), and p is the directed distance from the directrix to the vertex and from the vertex to the focus.<br>
Here is a problem like your problems 1 and 2....<br>
{{{x^2-4x+4y=0}}}<br>
Complete the square in x and put the equation in vertex form.<br>
{{{(x^2-4x+4)+4y-4 = 0}}}
{{{(x-2)^2+4y-4=0}}}
{{{4y-4=(x-2)^2}}}
{{{4(y-4)=(x-2)^2}}}
{{{y-4=(1/4)(x-2)^2}}}<br>
That is vertex form: h=2; k=4; 4p=4 so p=1.<br>
The vertex is (h,k) = (2,4).<br>
p=1, so the directed distance from the directrix to the vertex is 1, and the directed distance from the vertex to the focus is 1.  That makes the directrix y=3 and the focus (2,5).<br>
Your problem 5 is solved in a similar fashion.<br>
For problems 3 and 4, use the definitions of (h,k) and p to write the equations.<br>