Question

.

Answer 1

am i right? @satellite73

Answer 2

i think you have a typo there

Answer 3

it is true, you are right

Answer 4

now you are wrong, it is True

Answer 5

yw

Answer 6

focal width requires a google because i totally forget what that is
maybe someone else knows off the top of their head

Answer 7

oh wait i think the "focal length" of \(x=\frac{1}{4}y^2\) is just \(\frac{1}{4}\)

Answer 8

oh no that is wrong

Answer 9

\[x=\frac{1}{4}y^2\\
4x=y^2\] so the focal length is 4 i think

Answer 10

for your last question
\[-8x=y^2\\
4px=y^2\] not sure if \(p=2\) or \(p=-2\)
i think maybe \(p\) is always positive, but i would not bet on it