4x^2 = 7y How do they get the following for the focus and directix?

[Jurrasic]

Please don't leave out very many steps, please do give the formulas if possible please thank you:
How did they get these values?
Focus = (0,7/16)
directrix = -7/16

Tried to use p=1/4(a) which DID NOT work , that yields p = 1
THEN the focus would be (0,1) which it is not. Teacher said to use that formula but maybe that was for who knows what , because it's not working.

 

[boaz]

[itex]4x^2=7y \rightarrow 4x^2+0x+0=7y \rightarrow (4/7)x^2+0x+0=y[/itex]
thus, the focus is
[itex]
(-\frac{0}{2\cdot(\frac{4}{7})}, -\frac{0^2}{4\cdot(\frac{4}{7})}+0+\frac{1}{4\frac{4}{7}} ) = ( 0, \frac{7}{16} )
[/itex]
the same thing goes with the directrix.

hope it helped.

 

[eumyang]

^No offense, but I don't know how that post could have helped. It was hard for me to read.

Tried to use p=1/4(a) which DID NOT work , that yields p = 1
THEN the focus would be (0,1) which it is not. Teacher said to use that formula but maybe that was for who knows what , because it's not working.

You are not applying the formula correctly. One of the equations of the parabola is
x² = 4py,
but your equation is
4x² = 7y.
You have a coefficient for the x² term. So the first thing you must do is to divide both sides by 4.
[itex]x^2 = \frac{7}{4}y[/itex]
Now you can figure out the focus and directrix.