SOLUTION: The graph of the equation x^2 + y^2 + 8x − 10y − 40 = 0 is which of the following conic sections? A. parabola B. hyperbola C. ellipse D. circle

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The graph of the equation x^2 + y^2 + 8x − 10y − 40 = 0 is which of the following conic sections? A. parabola B. hyperbola C. ellipse D. circle      Log On


   

Question 974686: The graph of the equation x^2 + y^2 + 8x − 10y − 40 = 0 is which of the following conic sections?
A. parabola
B. hyperbola
C. ellipse
D. circle
Found 2 solutions by Fombitz, htmentor:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B+y%5E2+%2B+8x+%26%238722%3B+10y+%26%238722%3B+40+=+0
x%5E2%2B8x%2By%5E2-10y=40
x%5E2%2B8x%2B16%2By%5E2-10y%2B25=40%2B16%2B25
%28x%2B4%29%5E2%2B%28y-5%29%5E2=81
It's a circle centered at (-4,5) with a radius of 9.

Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
If we rearrange the equation, grouping the x and y terms, and move the constant term to the other side, we have:
x^2 + 8x + y^2 - 10y = 40
Complete the squares:
(1) (x+4)^2 = x^2 + 8x + 16
(2) (y-5)^2 = y^2 - 10y + 25
Since the constant terms add to 41, we need to subtract this from the LHS in order to write the equation as the sum of (1) and (2):
x^2 + 8x + y^2 - 10y -> (x+4)^2 + (y-5)^2 - 41 = 40 -> (x+4)^2 + (y-5)^2 = 81
This is the equation for a circle with center (-4,5), and radius sqrt(81) = 9
Graph is shown below.
Ans: D circle