Question

Will Medal and Fan ;-; Just Help me

Regan is trying to find the equation of a quadratic that has a focus of (–2, 5) and a directrix of y = 13. Describe to Regan your preferred method for deriving the equation. Make sure you use Regan's situation as a model to help her understand.

Answer 1

@naate

Answer 2

this is not nearly has hard as it looks
what is halfway between the horizontal line \(y=13\) and the point \((-2,5)\)?

Answer 3

which is the same question as "what is halfway between 5 and 13?" or even "what is the average of 5 and 13?"

Answer 4

9

Answer 5

i agree, so the vertex is this sucker is \((-2,9)\)

Answer 6

do you know whether this faces up or down?

Answer 7

um down?

Answer 8

ummm yes
the directrix is above the focus, so it opens down

Answer 9

equation will be \[(y-k)^2=4p(x-h)\] you know \(x=-2,k=5\) all that is left is \(p\)

Answer 10

so how do I get p?

Answer 11

to find it, \(p\) is the distance between the focus and the vertex, i.e. between \((-2,5)\) and \((-2,9)\) or equivalently the distance between the vertex \((-2,9)\) and the ditrectrix \(y=13\)

Answer 12

how far from 5 to 9?

Answer 13

4

Answer 14

so it's
(y−k)2=4p(x−h)
x=−2,k=5 p=4

Answer 15

yes, but make sure to put a minus sign in front of the \(4p\)since it opens down
let me know what you get

Answer 16

so it's
(y−k)2=4p(x−h)
x=−2,k=5 -4p

Answer 17

silly rabbit, replace the variables by the numbers !

Answer 18

;-; ohhhh so it's
so it's
(y−5)2=4(4)(-2−h)

Answer 19

\[(y−k)^2=4p(x−h)\] replace \(h\) by \(-2\) replace \(k\) by \(9\) and replace \(p\) by \(-4\)

Answer 20

\(k=9\) not \(5\)

Answer 21

and you need to replace \(h\) by \(-2\), not \(x\)

Answer 22

\[(y−9)=4(4)(-2−h)\]

Answer 23

closer, not yet

Answer 24

\[(y−5)2=4(4)(x−-2)\]

Answer 25

not \(-2-h\) put \(h=-2\) not \(x\) that is your variable

Answer 26

closer still, but \(p=-4\) because it opens down
also if you write \(--2\) your teacher will think you are daft
just make it \(+2\)

Answer 27

\[(y−5)2=4(-4)(x+2)\] did I do it right this time?

Answer 28

what would be even better is \[(y-9)^2=-16(x+2)\]

Answer 29

don't forget vertex was \((-2,9)\)

Answer 30

want to check it?

Answer 31

oh damn i made a huge mistake, sorry

Answer 32

djcnpasijdb ijdcijkdc j ;-; that is my mind right now I dislike math right now ;-; and yes it would be better I forgot all about the vertex ;-;

Answer 33

good thing i checked the answer

Answer 34

since it opens down the x term is squared, not the y term
try \[-16(y-9)=(x+2)^2\]

Answer 35

my mistake, sorry

Answer 36

is that the answer?

Answer 37

@satellite73

Answer 38

@clayton4christ ?

Answer 39

Sorry. I am not at all good at math

Answer 40

;-; its okay