Question

find the vertex, focus, directrix, and the length of the focal chord for y^2+6y-2x+13=0

Answer 1

change it to vertex form

Answer 2

how do you change it ?

Answer 3

y=-13(x-2)?

Answer 4

\[\large y^2+6y-2x+13=0\]

send y terms to the other side :

\[\large 2x-13= y^2+6y\]

Answer 5

complete the square for right hand side

Answer 6

knw how to complete the square ?

Answer 7

um dont you square root both sides ?

Answer 8

nope, we're trying to change it into below form :
\[\large \color{red}{x-h = a(y-k)^2}\]

Answer 9

so dont we add the 13 to both sides i am so sorry i honestly had a hard time converting into vertex my last packet :/

Answer 10

converting to vertex form is easy, all we need to know is the identity : \(\large \color{red}{a^2+2ab+b^2 = (a+b)^2}\)

Answer 11

\[\large 2x-13= y^2+6y \]

\[\large 2x-13= y^2+2(3)y \]

add 3^2 both sides :

\[\large 2x-13 + \color{green}{3^2}= y^2+2(3)y + \color{green}{3^2}\]

\[\large 2x-13 + \color{green}{3^2}= (y+3)^2\]

Answer 12

simplifying the left hand side gives you :

\[\large 2x-4= (y+3)^2\]

\[\large 2(x-2)= (y+3)^2\]

\[\large x-2= \dfrac{1}{2}(y+3)^2\]

Answer 13

Now we're ready to compute vertex/focus etc... using this form

Answer 14

compare above equation with the equation in second column :

Answer 15

before comparing, lets send the 2 to right side :

\[\large x = \dfrac{1}{2}(y+3)^2 + 2\]

Answer 16

now compare :)

vertex = \((h, k )\) = ?

Answer 17

(2,3)?

Answer 18

oh yes my bad

Answer 19

no its my bad ! you're right, I have misinterpreted the equation :o
let me fix it quick

Answer 20

compare below two :
\[\large \large x = \dfrac{1}{2}(y+3)^2 + 2 \]
\[\large \color{Red}{ x = a(y-k)^2 + h}\]

Answer 21

Vertex = \(\large (h, k) = (2, -3)\)

Answer 22

fine ? what else we need to find ?

Answer 23

the directirx and the length of the focal chord

Answer 24

lets find the directrix,
look at the table and can you tell me the equation for Directrix ? :)

Answer 25

you need to look at second column in the table..

Answer 26

y=k-c or x=h-c

Answer 27

our equation is in the form shown in the last column,
so the equation of directrix is \(\large x = h-c\)

Answer 28

we know \(\large h = 2\)
how to find \(\large c\) ?

Answer 29

use this from the table :

\[\large a = \dfrac{1}{4c}\]

Answer 30

compare the equations again, whats the value of \(a\) ?

Answer 31

1/2

Answer 32

yes !

\[\large \dfrac{1}{2} = \dfrac{1}{4c}\]

\[\large c = \dfrac{1}{2}\]

Answer 33

whats the equation of directrix ?

Answer 34

y=2-1/2

Answer 35

Yep! but nobody leaves it like that... simplify :)

Answer 36

hey wait, its `x = 2-1/2` right ?

Answer 37

oh yeah i got them mixed up

Answer 38

?

Answer 39

so what does that simplify to

Answer 40

directrix : \(\large x = \dfrac{3}{2}\)

Answer 41

what is the length of the focal chord?

Answer 42

memorize this formula :
length of focal chord = \(\large 4c\)

Answer 43

in our case \(c = \frac{1}{2}\), so focal chord = ?

Answer 44

2

Answer 45

yes ! and u may add above formula to the previous table...

Answer 46

oh and my bad i just need the focus for it and its done

Answer 47

oh we still need to find focus is it,
its easy, we just look at the last column in table :)

whats the focus formula ?

Answer 48

(h,k+c) or (h+c,k)

Answer 49

whats there in the right most last column ?

Answer 50

(h+c,k)?

Answer 51

Yep !

Answer 52

do u have all the values ? just plug them

Answer 53

(5/2,-3)

Answer 54

Perfect !

Answer 55

you will need to memorize the table somehow... atleast for the duration you're in conics module

Answer 56

do you know any good sites to be able to draw a graph for this?

Answer 57

yes below is best for graphing online
https://www.desmos.com/calculator

Answer 58

enter the equation

`y^2+6y-2x+13=0`

in left top textbox and press enter

Answer 59

cool thank you for taking this long with me i really appreciate it and you really helped me alot this summer :)

Answer 60

good to hear that :) yw :)

Answer 61

does the site work for any conic sections btw :o

Answer 62

yes it works for graphing all kinds of equations !

Answer 63

a bit more sophisticated online graphing tool :
http://www.geogebra.org/webstart/geogebra.html

Answer 64

do you know another website :P?