Question

anyone help me
another problem
2. Reduce each equation to standard form then find the coordinates of the center the foci, the end of the major axes and the ends of each latus rectum. Sketch the curve.

Given problem:
4xsquared + ysquared +8x -4y-8=0

Answer 1

Standard form:
\[\frac{(x+1)^{2}}{4} + \frac{(y-2)^{2}}{16} = 1\]
Center: (-1,2)
Foci: (-1,2+-2sqrt(3))
Major axis: (-1,2 +-4)
Minor axis: (-1 +-2, 2)

Answer 2

thankyou dumbcow for helping me but i want to know how you get these answer. im scared if our prof: ask me where i get my answer then i cant answer him

Answer 3

First group x's and y's together and move constants on right hand side
(4x^2 +8x) + (y^2-4y) = 8
Then use completing the square
Factor the 4 out of x's part first
4(x^2+2x) +(y^2-4y) = 8
4((x+1)^2-1) +(y-2)^2-4 = 8
distribute 4 back in
4(x+1)^2 -4 +(y-2)^2 -4 = 8
move constants to right side
4(x+1)^2 + (y-2)^2 = 16
Now divide by 16 to get 1 on right side
(x+1)^2 /4 + (y-2)^2 /16 = 1
Now you have it in standard form
The rest is pretty straightforward once you have it in this form
Use the attached reference sheet for ellipses to help you
hope this helps