Question

3x^2-30x-y+77=0
Find vertex , focus , directrix

Answer 1

for vertex i got 5,2

Answer 2

y-2=(x-5)^2 don't know the distance of the focus or directrix Hellpp

Answer 3

hmm

Answer 4

hold the mayo

Answer 5

okk

Answer 6

wondering how you got the vertex

Answer 7

but the focus and directrix are equidistant from the vertex

Answer 8

y-2=(x-5)^2

Answer 9

well... I see that.. how did you get it though? just off the graph?

Answer 10

or, are you supposed to get it just from the graph?

Answer 11

from the equation got x to one side and y to the other

Answer 12

\[3(x^2-10+25)=y-77\]

Answer 13

hmm 3 * 25 = 75 \(\ne 77\)

Answer 14

yea

Answer 15

\[3x^2-30+75=y-2\]

Answer 16

\[x^2-30+225=y-2\]

Answer 17

hmmm right... I see, you completed the square
right... ok well the vertex is correct

Answer 18

\[(x-5)^2=y-2\]

Answer 19

but i didn't get a the distance from 1/4p

Answer 20

hmmm actually ... lemme do it.... just to check

Answer 21

or maybe the 1/4p=1 and then the distance would be 1/4

Answer 22

\(\bf 3x^2-30x-y+77=0\implies 3x^2-30x+75+2=y
\\ \quad \\
3(x^2-10x+25)=y-2\implies 3(x-{\color{brown}{ 5}})^2=(y-{\color{blue}{ 2}})
\\ \quad \\
\begin{array}{llll}

(x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\
\end{array}

\qquad
\begin{array}{llll}
vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\
{\color{purple}{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\qquad thus
\\ \quad \\
3(x-{\color{brown}{ 5}})^2=(y-{\color{blue}{ 2}})\implies (x-{\color{brown}{ 5}})^2=\cfrac{1}{3}(y-{\color{blue}{ 2}})\qquad 4{\color{purple}{ p}}=\cfrac{1}{3}\implies {\color{purple}{ p}}=\cfrac{1}{12}
\\ \quad \\
directrix\implies y={\color{blue}{ 2}}+{\color{purple}{ p}}\qquad focus\implies ({\color{brown}{ 5}},{\color{blue}{ 2}}+{\color{purple}{ p}})\)

Answer 23

you kinda hosed the "3" in front of the binomial with the "x", otherwise, is fine

Answer 24

i thought dividing the 15 by 3 will eliminate the 3

Answer 25

isn't the 1/3 supposed to be with (x-h)^2

Answer 26

hmm nope, recall, it was the common factor

Answer 27

y-k=a(x-h)^2

Answer 28

ohh that

Answer 29

or x-h=a(y-k)^2

Answer 30

\(\large {\begin{array}{llll}
(y-{\color{blue}{ k}})^2=4{\color{purple}{ p}}(x-{\color{brown}{ h}}) \\
(x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\
\end{array}

\qquad
\begin{array}{llll}
vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\
{\color{purple}{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}
}\)

Answer 31

for the "focus form", used there, nope, the coefficient goes with the non-squared binomial, and the number gives us "4P"

where "P" is the distance from the vertex to the focus or directrix

Answer 32

so the 4p could be on any side because its just the distance?

Answer 33

but I know what you mean, you're referring to the "vertex form" of a parabola \(\bf y=a(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}\\
x=a(y-{\color{blue}{ k}})^2+{\color{brown}{ h}}\qquad\qquad vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\)

Answer 34

oh

Answer 35

thanks

Answer 36

yw

Answer 37

hmmm actually as a note
the focus and directrix are opposit to each other

so the "x" is squared, and has a positive coefficient
meaning the parabola opens upward

so the focus is "above" the vertex
and the direcrix is below it, each 1/12 units from the vertex

Answer 38

since I used above 2+p for both the directrix and focus

so the directrix, goes down from there or \(\Large y = 2-p\)
and the focus is just \(\Large (5, 2+p)\)

Answer 39

\(\bf 3(x-{\color{brown}{ 5}})^2=(y-{\color{blue}{ 2}})\implies (x-{\color{brown}{ 5}})^2=\cfrac{1}{3}(y-{\color{blue}{ 2}})
\\ \quad \\
4{\color{purple}{ p}}=\cfrac{1}{3}\implies {\color{purple}{ p}}=\cfrac{1}{12}
\\ \quad \\
directrix\implies y={\color{blue}{ 2}}-{\color{purple}{ p}}\qquad focus\implies ({\color{brown}{ 5}},{\color{blue}{ 2}}+{\color{purple}{ p}})\)