Question

Focal Width, Focus and directrix help.

Answer 1

\[-\frac{ 1 }{ 40 }x^2=y\]

Answer 2

\(\bf -\cfrac{ 1 }{ 40 }x^2=y\implies -\cfrac{ 1 }{ 40 }(x-0)^2=(y-0)
\\ \quad \\
(x-{\color{brown}{ h}})^2=4{\color{blue}{ p}}(y-{\color{brown}{ k}})\\ \quad \\\quad vertex\ ({\color{brown}{ h,k}})\quad {\color{blue}{ p}}=\textit{distance from the focus to vertex}\)

so.... if you were to simplify the equation to the "standard form" above... what would it look like?

Answer 3

-(x^2/160)(x-0)^2=(y-0)

Answer 4

what is the focal width? like whats the formula

Answer 5

-(x^2/160)(x-0)^2=(y-0) ?

Answer 6

that's the standard form you asked for

Answer 7

is the focal width 1/4(a)

Answer 8

well.. the idea is that you'd simplify it to the standard form

once you know where the vertex is and "p" distance.. you can pretty much get the rest

Answer 9

\(\large (x-{\color{brown}{ h}})^2=4{\color{blue}{ p}}(y-{\color{brown}{ k}})\leftarrow \textit{standard form}\)

Answer 10

(x-0)^2=4(P idk what p is)(Y-0)

Answer 11

well.... yes.. how about turning \(\bf -\cfrac{ 1 }{ 40 }(x-0)^2=(y-0)\) into \(\bf (x-{\color{brown}{ h}})^2=4{\color{blue}{ p}}(y-{\color{brown}{ k}})?\)

Answer 12

meaning that number present is the p? the -1/40 ?

Answer 13

well.. yes.. BUT when is next to the non-squared variable

Answer 14

well..is 4p really..not "p", "p" is just part of that number

Answer 15

(x-0)^2=4(-1/40)(y-0)

Answer 16

(x-0)^2=-1/10(y-0)

Answer 17

-1/10?

Answer 18

well... how about just multiplying -40 to both sides?

Answer 19

i multiplied the 4 by (-1/40)

Answer 20

what do you mean to both sides

Answer 21

\(\bf \cfrac{ (x-0)^2 }{ -40 }=(y-0)\implies (x-0)^2=-40(y-0)\) <---- see the vertex, see the "p" distance?

Answer 22

\(\bf -\cfrac{ 1 }{ 40 }x^2=y\implies -\cfrac{ 1 }{ 40 }(x-0)^2=(y-0)
\\ \quad \\
(x-{\color{brown}{ h}})^2=4{\color{blue}{ p}}(y-{\color{brown}{ k}})\\ \quad \\\quad vertex\ ({\color{brown}{ h,k}})\quad {\color{blue}{ p}}=\textit{distance from the focus to vertex}
\\ \quad \\
\cfrac{ (x-0)^2 }{ -40 }=(y-0)\implies (x-{\color{brown}{ 0}})^2=-40(y-{\color{brown}{ 0}})
\\ \quad \\
4{\color{blue}{ p}}=-40\implies {\color{blue}{ p}}=?\)

Answer 23

did you multiply -1/40 by (y-0)

Answer 24

p is -10

Answer 25

but im not understanding how we go there

Answer 26

hmm nope... if you use -1/40 that'd give you \(\bf \cfrac{ 1 }{ 160 }(x-0)^2=-\cfrac{(y-0)}{40}\)

Answer 27

ohhh wait i saw what you did you multiplied -40 to get rid of it

Answer 28

yeap

Answer 29

so the -10 is the directrix bc of y=0-10

Answer 30

focus (0,10)

Answer 31

yeap p = -10 so is a negative figure.

notice the "x" is the squared variable
that means the parabola is vertical

notice "p" is negative
that means the parabola is opening downward

notice the vertex, is the origin

so let's go from the vertex DOWNWARDS 10 units, and we'll find the focus point
the focal width is btw "4p"

and then to find the directrix... the directrix is the same "p" distance from the vertex
but in the opposite direction