Question

please check my work ^.^
medal + fan :]
if i got it wrong don't be too harsh, maths is not my best subject

Answer 1

@campbell_st @phi

Answer 2

I dont know if this is right or not, but please go back and recheck what the problem is saying, i think you may have made a mistake.

Answer 3

Well are you 100% sure about that? Cause lol I'm not just going to go back on an assumption. No offense but if you are not sure then I won't redo my work.

Answer 4

Im 100% sure :)

Answer 5

your solution is correct

Answer 6

Oh wow really @campbell_st ? o.O you sure? lol I trust ya but like I seriously can't believe I did something right?

Answer 7

Did you check it? becuase sometimes my nname deceives me, and other times im just plane wong, so i would listen to Campbell . Sorry!

Answer 8

the perpendicular distance from the directrix to the focus is 6 units

so the vertex is halfway between the focus and directrix on the line x = -5

so the vertex is at (-5, 2)

so half od 6 is 3

this is the focal length, a = 3

the version of the equation I use is

\[(x - h)^2 = 4a(y - k)\]

where (h, k) is the vertex and a is the focal length

so substittuting you get

\[(x + 5)^2 = 4 \times 3(y - 2)\]

now make y the subject

hope it helps

Answer 9

You are correct.

Answer 10

Thanks for that info, it does help! :]

Answer 11

@mathstudent55 thank you :]

Answer 12

I can't give both of you medals :[

Answer 13

\(\sqrt{(x + 5)^2 + (y -5)^2} = \sqrt{(y + 1)^2} \)

\((x + 5)^2 + (y -5)^2 = (y + 1)^2 \)

\((x + 5)^2 + \cancel{y^2} -10y + 25 = \cancel{y^2} + 2y + 1\)

\((x+5)^2 + 24 = 12y\)

\( 12y =(x+5)^2 + 24 \)

\(y = \dfrac{1}{12}(x + 5)^2 + 2\)

Answer 14

You're welcome.

Answer 15

Where did the problem go?
Keep the problem the way it was because this may help other people.