Question

Derive the equation of the parabola with a focus at (2, -1) and a directrix of y = - one half

Answer 1

okay so first ill put it in the formula and take the square root off

Answer 2

@haleyelizabeth2017

Answer 3

so (x-2)^2+(y+1)^2=(y+1/2)^2

Answer 4

right ?

Answer 5

one sec

Answer 6

yep

Answer 7

okay so then next i distribute y term binomials

Answer 8

Yes

Answer 9

What do you get when you simplify everything and isolate y?

Answer 10

so i got (x-2)^2=y^2+2y+1=y^2+y+1/4

Answer 11

you mean a + between the (x-2)^2 and y^2?

Answer 12

oops yes sorry

Answer 13

That is correct.

Answer 14

then subract y^2+y+1/4 from both sides

Answer 15

Nope, subtract the y^2+2y+1 from both sides

Answer 16

wait why

Answer 17

Because you want the (x-2)^2 to be on the opposite side

Answer 18

oh okay

Answer 19

so its (x-2)^2= -y-3/4

Answer 20

yes, so now, we can factor out -1 from -y and -3/4. Which gives us\[y+3/4=-(x-2)^2\] but then they will probably have you isolate the y leaving you with \[y=-(x-2)^2-3/4\]

Answer 21

okay got it one last one

Answer 22

Derive the equation of the parabola with a focus at (-2, 4) and a directrix of y = 6. Put the equation in standard form

Answer 23

i didnt get the right answer

Answer 24

haleyelizabeth2017

Answer 25

@haleyelizabeth2017

Answer 26

So sorry!

Answer 27

lol no its fine

Answer 28

what did you get when you got this one?

Answer 29

well i got 1/20(x+2)^2-1 but thats incorrect

Answer 30

How'd you get that?

Answer 31

I got -1/4(x+2)^2+5

Answer 32

can you explain one last time how you got that PLEEEEAAAASSSEEE

Answer 33

\[(x+2)^2+(y-4)^2=(y-6)^2\]\[(x+2)^2+y^2-8y+16=y^2-12y+36\]\[(x+2)^2=-4y+20\]\[(x+2)^2=-4(y-5)\]\[y-5=-1/4(x+2)^2\]\[y=-1/4(x+2)^2+5\]

Answer 34

okay thank you so much

Answer 35

No problem