Question

HELP ASAP

Answer 1

what is the question though

Answer 2

Graph each pair of parametric equations.

Answer 3

sigh the fastest way is a calculator, if it is allowed, but other than that, this is out of my range

Answer 4

how about any of these:

Answer 5

@DanJS @karim728 @jackthegreatest

Answer 6

I need help with those 3 please

Answer 7

The main idea of the first one is, remember the distance from any point on a parabola to the focus is the same as the distance from that point to the directrix...

Answer 8

@jackthegreatest

Answer 9

@jackthegreatest

Answer 10

@mathstudent55

Answer 11

@jackthegreatest

Answer 12

@sweetburger

Answer 13

For the first parabola, you just have to remember that the distance between any point on the parabola and the focus is the same distance from that point on the parabola to the directrix.

If you call the arbitrary point on the parabola (x,y),
Recall the distance formula \[d=\sqrt{(x-x _{0})^2+(y-y _{0})^2}\]

Answer 14

Distance from point (x,y) on the parabola, to the focus point (0, -2)
\[\sqrt{(x-0)^2+(y-(-2))^2}\]

Distance from point (x,y) on the parabola, to the directrix y=2
\[\left| y-2 \right|\]

The property of parabola is that those are both the same

\[\large \sqrt{x^2+(y+2)^2}=\left| y-2 \right|\]

Answer 15

You can simplify that down to the standard form y=ax^2 + bx + c

squaring an absolute value can remove the absolute value, squaring a negative is positive anyway

\[x^2 + (y+2)^2=(y-2)^2\]
\[x^2 + y^2+4y+4=y^2-4y +4\]
\[x^2=-8y\]
\[\large y=\frac{ -1 }{ 8 }x^2\]

that is it

Answer 16

@DanJS @inkyvoyd @sweetburger