Question

Please this is an emergency! Can someone give me the vertex, focus, and directrix of this parabola?

Answer 1

@jim_thompson5910 @Lord_Box @zepdrix @Zarkon

Answer 2

First add 48 to both sides

\[\Large 12y = (x-1)^2 - 48\]
\[\Large 12y+48 = (x-1)^2 - 48+48\]
\[\Large 12y+48 = (x-1)^2\]

Answer 3

Then factor out 12 from the left side

\[\Large 12y+48 = (x-1)^2\]
\[\Large 12(y+4) = (x-1)^2\]

Answer 4

but what would the directrix and vertex and foci be?

Answer 5

\[\Large 12(y+4) = (x-1)^2\]
is the same as
\[\Large 4*{\color{red}{3}}(y-({\color{blue}{-4}})) = (x-{\color{green}{1}})^2\]
I rewrote the equation into the form `4p(y-k) = (x-h)^2` where in this case
\(\Large {\color{red}{p = 3}}\)
\(\Large {\color{blue}{k = -4}}\)
\(\Large {\color{green}{h = 1}}\)


see this article
http://www.mathwords.com/f/focus_parabola.htm

Answer 6

how do you get the directrix foci and vertex?

Answer 7

the vertex in general is (h,k)

Answer 8

for parabolas that open up, the focus is at (h,k+p)
for parabolas that open up, the directrix is y = k-p


for parabolas that open down, the focus is at (h,k-p)
for parabolas that open down, the directrix is y = k+p


p = distance from vertex to focus (also distance from vertex to directrix)

Answer 9

Thanks!

Answer 10

Sure thing