SOLUTION: it says, "Write each equation in standard form?" ellipse: 96(x+6)^2 +24(y-4)^2 = 96 hyperbola: 9(x+1)^2 - 16(y+12)^2 =144 parabola: x^2 + 12x + 36 = -8y + 8 i am soo lost!

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: it says, "Write each equation in standard form?" ellipse: 96(x+6)^2 +24(y-4)^2 = 96 hyperbola: 9(x+1)^2 - 16(y+12)^2 =144 parabola: x^2 + 12x + 36 = -8y + 8 i am soo lost!      Log On


   



Question 665083: it says, "Write each equation in standard form?"
ellipse: 96(x+6)^2 +24(y-4)^2 = 96
hyperbola: 9(x+1)^2 - 16(y+12)^2 =144
parabola: x^2 + 12x + 36 = -8y + 8
i am soo lost!

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,
Need to know Your Standard forms: See below
ellipse: 96(x+6)^2 +24(y-4)^2 = 96 ⇒ %28x%2B6%29%5E2%2F1+%2B%28y-4%29%5E2%2F4+=+1
hyperbola: 9(x+1)^2 - 16(y+12)^2 =144 ⇒ 9%28x%2B1%29%5E2%2F16+-+16%28y%2B12%29%5E2%2F9=1
parabola: x^2 + 12x + 36 = -8y + 8 ⇒ %28x%2B6%29%5E2=+-8%28y-1%29

Standard Form of an Equation of an Ellipse is %28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1+
where Pt(h,k) is the center. (a variable positioned to correspond with major axis)
a and b are the respective vertices distances from center
Standard Form of an Equation of an Hyperbola opening right and left is:
%28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 with C(h,k) and vertices 'a' units right and left of center,

the vertex form of a Parabola opening up(a>0) or down(a<0), y=a%28x-h%29%5E2+%2Bk
where(h,k) is the vertex and x = h is the Line of Symmetry
The standard form is %28x+-h%29%5E2+=+4p%28y+-k%29, where the focus is (h,k + p)