Question

I have a question about math derivatives...

Answer 1

I actually have 5 questions on the same thing, just different setups. But how would I go about solving them? .-.
1. Derive the equation of the parabola with a focus at (−5, −5) and a directrix of y = 7
2. Derive the equation of the parabola with a focus at (6, 2) and a directrix of y = 1
3. Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = -1.
4. Derive the equation of the parabola with a focus at (0, −4) and a directrix of y = 4
5. Derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. Put the equation in standard form.

Answer 2

If I can get assistance through one, I'm sure I can do the rest on my own (maybe)

Answer 3

The equation of a parabola with focus in (h,k+p) and directrix y=k-p, is,
\[(x-h)^2=4p(y-k)\]In the first case,
\[h=-5\\ k+p=-5\\y=k-p=7\]Combining the last two relations, k=1,p=-6. So the equation of the parabola is,
\[(x+5)^2=-24(y-1)\]

Answer 4

I understand a bit but not really. That also doesn't match any of the answer choices I have.. (thanks for replying)

Answer 5

I can only assume that it's actually f(x) = −1/24(x + 5)2 + 1 (which is what I chose in the beginning)

Answer 6

You can expand the formula to obtain the standard form
\[x^2+25+10x=-24y+24\Rightarrow y=-\frac{1}{24}(x^2+10x+1)\]

Answer 7

I think the form you cited,
\[y=-\frac{1}{24}(x+5)^2+1\]
is the vertex form, but you can deduce from the one I gave you, simply remember y=f(x) and then,
\[(x+5)^2=-24(f(x)-1)\Rightarrow -\frac{1}{24}(x+5)^2=f(x)-1\Rightarrow\\\Rightarrow f(x)=-\frac{1}{24}(x+5)^2+1\]
Do you understand it?

Answer 8

Kind of.... Unless you're busy, could you help me work through the rest? (it's only 5) or at least the second one. I understand a little how you transfer it, but not how you got it to begin with

Answer 9

I only have one question, question 5 that's in standard form

Answer 10

Oh, then was my mistake! I think they need to be expressed all in standard form, sorry.

Answer 11

all of the answer choices for each question have the same layout, though. it just looks really complex to me

Answer 12

Wll, with the second one, and all the others, you must follow this steps.

(1) Identify the focus,
\[ (h,k+p) =(6,2)\Rightarrow h=6\ \ \text{and} \ \ k+p=2 \](2) Identify the directrix,
\[y=k-p=1\](3)Find the valus k and p solving the system you get before,
\[k+p=2\\
k-p=1\]Solving this, you have k=3/2 and p=1/2.(4) Susbtitute the obtained values in the equation,
\[(x-h)^2=4p(y-k)\]In this case,
\[(x-6)^2=4\cdot\frac{1}{2}\cdot\left(y-\frac{3}{2} \right)\]Solving for y,
\[(x-6)^2=2\cdot\left(y-\frac{3}{2} \right)\Rightarrow \frac{1}{2}(x-6)^2+\frac{3}{2}=y\]

Answer 13

Do you understand the four steps?

Answer 14

And how to apply to the rest?

Answer 15

I understood it up until step 4.
working with question 3, according to your steps..
step 1: H= -5, K+P = 5?
step 2: y=k−p=-1
step 3: k+p = 5, k-p = -1

Answer 16

right?

Answer 17

Perfect. Now solve the system,
\[k+p=5\\
k-p=-1\]Simply, adding the equations,
\[(k+p)+(k-p)=5-1\Rightarrow 2k=4\Rightarrow k=2\]Then susbtitute this value in one of the equations above, for example,
\[k+p=5\Rightarrow2+p=5\Rightarrow p=3\]
Do you understand it?

Answer 18

I understand it goes from 5-1... what I don't see is the "2k=4" :O Unless it's just an obvious thing... like 3k would equal 6. (looking at it, 5-1=4 so is that how you got it?)
K = 2, okay.. (because that's half of 4?)

Answer 19

I'm trying to make sure I know this 100%, lol, sorry. I have a call right after this and I'm going to be questioned on it

Answer 20

You sum the part from the left,
\[(k+p)+(k-p)=k+k+p-p=2k+0=2k\]
Do you understand this step?

Answer 21

And this sum always gives 2k the only number that changes is the part of the right (in this case 5-1=4)

Answer 22

yep, understand that! Wouldn't that be the same for each question, though? Since there's only one set of "k+p, k-p" for each question

Answer 23

Exact!

Answer 24

haha yeah :D

Answer 25

so, taking that k+p or k-p determines that "2k=__"

Answer 26

Then only part that changes is the right member. Well, I have to go, but I give you the answer I obtained in order you can check it,
(3) \[y=\frac{1}{12}(x+5)^2+2\]
(4) \[y=-\frac{1}{16}(x)^2\]
(5) \[y=-\frac{1}{4}(x+2)^2+5\]
This last, in standard form is,
\[y=-\frac{x^2}{4}-\frac{x}{2}+5\]
I hope they are ok. And good luck!!

Answer 27

Yes, 2k=5-1, that is 2k=4, so solving this,
\[2k=4\Rightarrow k=4/2=2\]

Answer 28

Darn. Thanks a lot, though!

Answer 29

;) Sure here a lot of people will help if you need something more.

Answer 30

I'll try to work on 4 and post my work here. how long will you be gone? #5 is the only different problem

Answer 31

question 4:
step 1: H=0, K+P = -4
step 2: y=k−p=-4
step 3: k+p = -4, k-p = 4

Answer 32

it is the locus of a point s.t its distance from a fixed point called focus=distance from a fixed line called directrix.