Question

y = 5x^2

conic sections parabola. have to find focus, directrix, and diameter. Answers are F(0,1/20) for focus, -1/20 for directrix, and 1/5 for diameter. please explain how they got these answers :)

Answer 1

I know the setup is \[x ^{2}=20y\] but how are they getting the 1/4p times 5 to get that focus?

Answer 2

\(\bf y=5x^2\implies (x-h)^2=4p(y-k)\qquad so\\ \quad \\
y=5x^2\implies (y-0)=5(x-0)^2\)

(h, k) = vertex of the parabola
p = distance from the vertex to the focus, or directrix, since both are equidistant to the vertex

Answer 3

i see but the formulas my book gives me are x^2=4py and y^2=4px

Answer 4

well... yes, those equations are correct for a parabola whose vertex is the origin

Answer 5

heres the example they give that says should help solve this problem but to me its too simple to see where they they got those values and not enough written equations.

Answer 6

look at example 4

Answer 7

ok... anything wrong with that?

Answer 8

it makes since but this problem im doing its alittle different so im not sure how they got the 1/20 for p

Answer 9

1/2 you mean

Answer 10

see heres the books answer to this problem. Number 15

Answer 11

i know when you switch the formula from the original \[y = 5x^{2}\] you put it in the form \[x^{2} = 4py\] so that means it would be \[x^{2} = 20y\]

Answer 12

\(\bf y=5x^2\implies (x-h)^2=4p(y-k)\qquad so\\ \quad \\
y=5x^2\implies (y-0)=5(x-0)^2\implies \cfrac{1}{5}(y-0)=(x-0)^2\\ \quad \\\implies (x-0)^2=\cfrac{1}{5}(y-0)\\ \quad \\
\begin{array}{llll}
(x-h)^2=\color{red}{4p}(y-k)\\
(x-0)^2=\color{red}{\cfrac{1}{5}}(y-0)
\end{array}\qquad\implies 4p= \cfrac{1}{5} \implies p=?\)

Answer 13

so is my logic wrong?

Answer 14

sadly we didnt get taught that formula for the vertex in this Trignomentry college course

Answer 15

hmmm seems so.... once you put it in the form like above.... you can see what 4P equates to

Answer 16

so to get that 1/5 did you just divide it out from the \[x^{2}\]

Answer 17

you'd isolate the squared term, yes

Answer 18

\(\bf x^2=4py\implies (x-0)^2=4p(y-0)\implies (x-h)^2=4p(y-k)\\ \quad \\
y^2=4px\implies (y-0)^2=4p(x-0)\implies (y-k)^2=4p(x-h)\)

these two above, presume the vertex is at ( 0, 0) but the general forms are the ones with the (h, k) variables

Answer 19

ok, so how do you solve for the diameter to get that 1/5 or is it just assumed?

Answer 20

hmmm not sure on that one... since parabolas do not have diameters, circles do

Answer 21

yea i know they call this a focal diameter which makes up the latus rectum

Answer 22

thats why im confused :(

Answer 23

there was an example on yahoo answers that got 1/4p thats why i was lost :( how they got that

Answer 24

hmmm focal width... I see... focal width is 4P so 4 times P, \(\bf p = \cfrac{1}{20}\qquad \textit{focal width}\quad 4\cdot \cfrac{1}{20}\implies \cfrac{1}{5}\)

Answer 25

i got another problem like this after this one so what do you recommend on approaching it because it is very similar

Answer 26

same I'd think, change it to the general form first

Answer 27

ok, so this one says x=-8y^2

Answer 28

well, isolate the squared term first

Answer 29

i made it -1/8x because variables by them selves divided are essentially 1 correct

Answer 30

then \(\bf 4p=-\cfrac{1}{8}\)

Answer 31

\[y^{2} = -1/8 x\]

Answer 32

1/-32

Answer 33

yeap

Answer 34

yay i got them both right :) thank you!

Answer 35

yw