Last flipping question... conics!?

Question

find the focus/directrix of x=2y^2-3y-2. I just need help putting it into standard form, the 1st 2 is throwing me off. Help?

I need standard form of a parabola, which is y^2=4px.

The problem is that first two, I can't figure out how to complete the square with it... I'm probably over thinking it, I know...

Heh, I didn't even think of dividing all terms by 2 right away. I knew I was over thinking it... thanks!

Answer 1

Just arrange the terms!

Here it is:

x = 2y² - 3y - 2

½x = y² - (3/2)y - 1

½x + 1 = y² - (3/2)y

Adding (½*3/2)² ehich is (9/16) to both sides:

½x + 25/16 = y² - (3/2)y + 9/16

½x + 25/16 = [y-(3/4)]²

Put X = ½x + 25/16

and

Y = y-(3/4)

Now you get the form
Y² = X
which is a standard form with a = ¼
________________ ________________ ______________

Focus is (a,0)

Since the origin has been shifted we have further calculations:
X=a

½x + 25/16 = ¼

½x = -21/16
x = -21/8


Y = 0

y-(3/4) = 0

y = 3/4

Thus Focus ≡ (-21/8 , 3/4)

Directrix :
X = -a

½x + 25/16 = -¼

½x = -29/16
x = -29/8
or
x+29/8=0

Answer 2

x = 2y^2 - 3y - 2
x = 2(y^2 - (3/2)y - 1)
x = 2(y^2 - (3/2)y + 9/16 - 9/16 - 1)
x = 2(y - 3/4)^2y - 9/8 - 2
(x + 25/2) = 2(y - 3/2)^2y
1/4a = 2
a = 1/8
vertex is at (-13/2,3/2)
1/8 - 13/2 = 1/8 - 52/8 = -51/8
focus is at (- 51/8)
directrix is x = - 53/8

Answer 3

x = (2y + 1)(y - 2)

If x = 0 then,
y = -1/2 or y = 2

Answer 4

idk?? go to webmath.com (does it for you :] )