I've already done these problems but I want to make sure that I'm right. Can someone post what you get? 1. Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = -1. 2. Derive the equation of the parabola with a focus at (−7, 5) and a directrix of y = −11. 3. Derive the equation of the parabola with a focus at (0, 1) and a directrix of y = −1. 4. Derive the equation of the parabola with a focus at (2, −1) and a directrix of y = −1/2. 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.
I don't wanna do all of them in case you're being sneaky and just looking for answers lol XD I'll show you what I got for a couple of them though :) For 1. \(\Large\rm y=\frac{1}{12}(x+5)^2+2\) For 3. \(\Large\rm y=\frac{1}{4}x^2\) Getting same results or no? :o
for #3 i got 4x^2
Hmm maybe I made a boo boo, lemme check my work.
\[\Large\rm 4p(y-k)=(x-h)^2\]Given: \(\Large\rm F:(0,1)\) \(\Large\rm D:y=-1\) The p value is half of the distance from the focus to the directrix. From -1 to 1 is a distance of 2. Half of that is 1. So \(\Large\rm p=1\). Our parabola is opening upward because the focus is above the directrix. So no negative sign or anything. The vertex will be located just under the focus, a distance of p. \(\Large\rm V:(0,1-1)=(0,0)\) So our equation becomes:\[\Large\rm 4p(y-k)=(x-h)^2\]\[\Large\rm 4\cdot1(y-0)=(x-0)^2\]\[\Large\rm y=\frac{1}{4}x^2\]Hmm I think that's right.. ya? :o maybe? :o
yeah you're right...i forgot about the invisible 1 in front of \[(x-0)^2\]
What about the rest?
maybe you tell me what you got and i can check? ;) So i don't suspect that you're fishing for answers haha XD
haha okay for #2 i got f(x)=−1/32(x−7)^2−3 #4 i got f(x)=−(x−2)^2+3/4 #5 i got f(x)= (1/4)x^2−x+5 @zepdrix
Hmm they all look just a litttttle bit off :o let's check out #2 a sec
\(\Large\rm F:(-7,5)\) \(\Large\rm D:y=-11\) The distance from 5 to -11 is 16. So then, \(\Large\rm p=8\) So our vertex is located at, \(\Large\rm V:(-7,5-8)=(-7,-3)\) And again, we're opening upwards since our Focus is above our Directrix. \(\Large\rm 4\cdot8(y--3)=(x--7)^2\)
\(\Large\rm 4\cdot8(y+3)=(x+7)^2\) \(\Large\rm y=\frac{1}{32}(x+7)^2-3\) something like that, ya?
hmm, i don't see where i went wrong, must've made mistake in the signs some where
Did you get the same vertex as me? (-7,-3) ?
as i'm redoing my work , yes
hehe
lol, could you help me with the others?
#4 \(\Large\rm F:(2,-1)\) \(\Large\rm D:y=-\frac{1}{2}\) The distance between our -1 and -1/2 is 1/2. So our p is half of that, \(\Large\rm p=\frac{1}{4}\) Our vertex is located at ummm, \(\Large\rm V:\left(2,-1--\frac{1}{4}\right)=\left(2,-\frac{3}{4}\right)\) Our Focus is located LOWER than our directrix, so we'll be opening downward. (I guess that means I should have just called the p value -1/4, that's what I was using in my calculations.) \(\Large\rm 4\cdot\left(-\frac{1}{4}\right)\left(y--\frac{3}{4}\right)=(x-2)^2\)
\[\Large\rm y=-(x-2)^2-\frac{3}{4}\]
lol, my work is all correct, i just accidentally put an addition sign in front of 3/4, just careless errors haha
hehe :D
I didn't check number 5 yet, let's seeeee...
#5 \(\Large\rm F:(-2,4)\) \(\Large\rm D:y=6\) Focus is below the directrix, so our parabola is opening downward. So we want a negative p value. The distance from 4 to 6 is 2. Our p value is half of that. \(\Large\rm p=-1\) \(\Large\rm V:(-2,6-1)=(-2,5)\) Then, \(\Large\rm 4(-1)(y-5)=(x--2)^2\)
Expanding out the right side gives us,\[\Large\rm -4(y-5)=x^2+4x+4\]
Dividing by -4,\[\Large\rm y-5=-\frac{1}{4}x^2-x-1\]
Adding 5, \(\Large\rm y=-\frac{1}{4}x^2-x+4\) something like that ya? :o
I would assume the lil boo boo you made was, when you divided by 4, you probably wrote 4/4 as 0.
you're so right, it's like you're looking at my paper lol
haha XD just a common mistake i see
That's what they all were really...thanks a bunch!
yay team \c:/
haha! have a good day, zepdrix (:
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