Question

How to find the focus and directrix of a parabola?

Answer 1

I know about the equation \(\rm y^2 = 4ax\)

Answer 2

*locus

Answer 3

So, can we solve for the focus using the above?

Answer 4

No, it's focus.

Answer 5

for y^2 = 4ax

(a,0) is the focus

x+a = 0 is the directrix..

Answer 6

you may have to use defintion of parabola if you want a better understanding..

Answer 7

Yes.

Okay, so if I am given the function \(\rm y = -3x^2 + 2\), how will I find the focus?

Answer 8

Yes... the above has the vertex form as:\[\rm y = -3(x - 0)^2 + 2\]Vertex is (0,2).

Answer 9

y - 2 = -3x^2
we knoiw, in x^2 = 4ay

(0,a) is the focus
and y = -a the directrix

Answer 10

Wait... \(\rm y^2 = 4ax\)

Answer 11

if we try and convert this into desired form, we see

x^2 = 4 ( -1/12) (y-2)

had that -2 not been there, we'd have had

focus = (0,-1/12) and directrix y = 1/12

but there's a 2 subtracted from y..

Answer 12

Can you please teach me?

Answer 13

ohh..arent you getting it /.?

Answer 14

No. :(

Answer 15

\[\rm y - 2 = -3x^2\]I can get it till here.

Answer 16

But, isn't the equation \(\rm y^2 = 4ax\)?

Answer 17

keep in mind
for x^2 = 4ay ,
a is the focus and and directrix as usual ..

Answer 18

you have to compare
y-2 = 3x^2 with y^2 = 4ax

Answer 19

sorry,,not y^2 = 4ax,,but with x^2 = 4ay..

Answer 20

are you getting confused?

Answer 21

Isn't the equation supposed to be \(\rm y^2 = 4ax\) instead of \(\rm x^2 = 4ay\)?

Answer 22

am sorry am not a very good teacher..

Answer 23

No worries. :)