SOLUTION: Use 4x^2 - y^2 - 8x - 4y + 16 = 0 to do the following. Write the equation in standard form. I got (x-1)^2/2 - (y-2)^2/8 = 1 Locate the centre and vertices of this curve.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Use 4x^2 - y^2 - 8x - 4y + 16 = 0 to do the following. Write the equation in standard form. I got (x-1)^2/2 - (y-2)^2/8 = 1 Locate the centre and vertices of this curve.       Log On


   

Question 437895: Use 4x^2 - y^2 - 8x - 4y + 16 = 0 to do the following.
Write the equation in standard form.
I got (x-1)^2/2 - (y-2)^2/8 = 1
Locate the centre and vertices of this curve.
State the domain and range of this curve.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi, 

 4x^2 - y^2 - 8x - 4y + 16 = 0   |Multiplying thru by -1

  y^2 + 4y - 4x^2 + 8x - 16 = 0

 (y+2)^2 -4  - 4[(x-1)^2 -1]  - 16 = 0

 (y+2)^2 -4  - 4(x-1)^2 +4  - 16 = 0

 (y+2)^2 - 4(x-1)^2 = 16

 %28y%2B2%29%5E2%2F16+-++%28x-1%29%5E2%2F4+=+1 C(1,-2) Vertices: (1,-6) and (1,2)

Domain: and Range: {y  ≤ -6 , y ≥ 2}