Question

57

plz help...

find the vertex, focus, and directrix of the parabola and sketch its graph

x^2 + 4x + 6y - 2= 0

Answer 1

u do not have to graph it

Answer 2

a good explanation is here
http://www.purplemath.com/modules/parabola.htm

Answer 3

i already know all these things but i do not know how to apply it to this problem

Answer 4

answer is here
http://www.wolframalpha.com/input/?i=focus+x^2+%2B+4x+%2B+6y+-+2%3D+0
but i think to get there we have to rewrite this thing

Answer 5

can u plz show me how they got the focus,vertex and directrix instead of just having the answers

Answer 6

ok lets see if we can write it in the form
\[4p(y-k)^2=(x-h)^2\]

Answer 7

then we can read off the answer. so the next part is just algebra

Answer 8

\[x^2 + 4x + 6y - 2= 0\]
\[6y-2=-x^2-4x\] now complete the square on the right to get
\[6y-2=-(x^2+4x)\]
\[6y-2=-(x+2)^2+4\]

Answer 9

does that part make sense?

Answer 10

the equation is actually (x-h)^2 = 4p (y-k)

Answer 11

equality is reflexive, so we are free to write it either way

Answer 12

oh damn i see i made a typo!

Answer 13

that is what is meant

Answer 14

sorry about that
it should be
\[4p(y-k)=(x-h)^2\]

Answer 15

ok so far we have
\[6y-2=-(x+2)^2+4\] is that part ok? only two more steps

Answer 16

\[6y-6=-(x+2)^2\]
\[6(y-1)=-(x+2)^2\]
\[-6(y-1)=(x+2)^2\] now we can read off the answer

Answer 17

where did u get the +4 at the end from?

Answer 18

ok lets go over that slowly
focusing only on the right hand side, we had
\[-x^2-4x\]

Answer 19

factor out the \(-1\) get \(-(x^2+4x)\) and now we complete the square

Answer 20

half of 4 is 2, so we write \(-(x-2)^2+4\) and the reason for the \(+4\) is that
\[-(x+2)^2=-x^2-4x-4\] so we have actually subtracted 4 and so now we must add it back, otherwise it is no longer what we started with

Answer 21

got it ok so what is the next step?

Answer 22

for that matter we could have written
\[6y-2=-(x^2+4x)\] and then
\[6y-2-4=-(x+2)^2\] because if we subtract 4 from the right, we have to subtract 4 form the left
either way we end up with
\[6y-6=-(x+2)^2\]

Answer 23

that is actually either but i understand it both ways

Answer 24

actually easier*

Answer 25

the we get to the form we want, namely
\[4p(y-k)=(x-h)^2\] by multiplying both sides by \(-1\) to get
\[-6(y-1)^2=(x+2)^2\]

Answer 26

yeah probably easier since we have an equation. the important thing to remember is that because of the minus sign out front you are subtracting 4, not adding it

Answer 27

now we should be able to read the answer from the form we have
we know it is a parabola that opens down. we knew that from the start

Answer 28

vertex is \((h,k)\) in this case it is \((-2,1)\)

Answer 29

is that another typo when u say -6 (y-1) ^2 above

Answer 30

yeah something is the matter with me

Answer 31

this is the form we have and it is what we want
\[-6(y-1)=(x+2)^2\]

Answer 32

so we can read off the vertex right?

Answer 33

then we know that it is a parabola opening down with vertex at \((-2,1)\) now we can find the focus

Answer 34

\[4p=-6\implies p=-\frac{3}{2}\] we go down \(-\frac{3}{2}\) units from \((-2,1)\) and end up at \((-2,-\frac{1}{2})\) so that is the focus

Answer 35

thanks so much
i got it from this point on

Answer 36

and finally the directrix is \(\frac{3}{2}\) units above \((-2,1)\) namely
\(y=1+\frac{3}{2}=\frac{5}{2}\)

Answer 37

yw

Answer 38

and sorry about the typos

Answer 39

it is ok