The first one is an ellipse centered at (x,y)=(-1,-3) with semi-major axis of √(20), semi-minor axis of √(20/3) .
The second one is a hyperbola:
The third is a parabola:
Hello, so basically I have previously taken applied calculus about 2 years ago. Friend is now taking precalc and asked for help. Needless to say..ive forgotten much of what I learned. So before I go on helping..have some problems I am attempting that I want to check myself on. So I will probably update this with various questions to check myself if that's ok. Thanks
First up, putting equation into standard form for a conic. (are all these ellipses?)
3x^2-y^2+6x+6y - 10 = 0
So complete the square I ended up with
3(x+1)^2-(y+3)^2 = 20
putting in standard form by dividing by 20 I came up with
[3(X+1)^2]/20-[(y+3)^2]/20 = 1
is this answer correct? Thanks
4x^2+9y^2+24x+18y+9 = 0
4(x+3)^2=9(y+1)^2 = 1******-9 + (1^2)+(3^2) = 1
since already = 1..is in standard form?
last question for standard form
x^2+10x = -2y+1
x^2+10x+2y = 1
(x+5)^2+2y = 26
[((x+5)^2)/26]+(2y/26) = 1
1) This is not an ellipse.
You can tell what a conic will be by looking at A and B in the equation Ax^2 + By^2 +.... = 0
A = B --> circle
A or B = 0 --> parabola
A, B have same sign --> ellipse
A, B have different sign --> hyperbola.
We have a HYPERBOLA (or a typo...)
2) Let's clear up 1) before proceeding
So new #1.
4(x+2)^2+6(y-3)^2 = 24
came up with ((x+2)^2/6) + ((y-3)^2)/4 = 1
Center (-2,3) (being reversed because standard form is (x-h) and (y-k) correct?
c^2 = 20. (foci distance)
vertices being (-2, 3+sqrt(6)) and (-2, 3-sqrt(6))
do I have these correct?
still looking for foci and minor vertices
so the semi vertices are b, in this case 2.
When determining the vertices, how do you determine when when you should move on the x or y axis?
if going with wider than taller (a^2 going under x and b^2 under y) you just associate it this way? since b is under the y section, you move on the y axis? So if it was taller than wider, and B^2 was under x, then b^2 moves along the x axis?
alright, lets see if I can do better. Moving on
16x^2-4y^2 = 64. So Hyperbola
(x^2/4)-(y^2/16) = 1
C^2 = 4+16 = 20
So Center is (0,0)
since x-y then vertices (2,0)(-2,0)
Foci (0, sqrt(20)) (0,-sqrt(20))
asym = y= +/- 2x
I have this one correct?
(y-2)^2 = 10(x+1) = parabola
4p = 10 p = 5/2
Focus (-1+5/2, 2)
Dirextrix x= -1-5/2
Axis of Sym y=2
Right now just want to check my work. I dont have an answer sheet for these problems, so just making sure im doing right
trouble factoring this equation..really stumped..can't get come out even :/
3x^2-y^2+6x+6y-10 = 0
3x^2+6x-y^2+6y = 10
3(x^2+2x)-(y^2+6y) = 10
3(x^2+2x+1) - (y^2+6y+9) = 10+9
3(x+1)^2=3(1) - (y+3)^2 = 19+3..but 3 doesn't divide evenly into 22.. like wise..if I just subtract out the 1..doesn't divide evenly into 20 :/