SOLUTION: the graph of the equation below is an ellipse. How long is the major axis of the ellipse given by the equation? (x+7)^2 5^2 + (y-4)^2 8^2= 1

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: the graph of the equation below is an ellipse. How long is the major axis of the ellipse given by the equation? (x+7)^2 5^2 + (y-4)^2 8^2= 1      Log On


   

Question 723537: the graph of the equation below is an ellipse. How long is the major axis of the ellipse given by the equation? (x+7)^2 5^2 + (y-4)^2 8^2= 1
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If the equation is really %28x%2B7%29%5E2+%2A5%5E2+%2B+%28y-4%29%5E2%2A+8%5E2=+1 the major and minor axes are not very long.
graph%28300%2C300%2C-8%2C1%2C-2%2C8%2C25%28x%2B7%29%5E2%2B64%28y-4%29%5E2%3C=1%29
When y=4 the second term is %284-4%29%5E2%2A+8%5E2=0%5E2%2A+8%5E2=0%2A+8%5E2=0
and you get the extreme values for x:
%28x%2B7%29%5E2+%2A5%5E2+=+1 --> %28x%2B7%29%5E2+=+1%2F5%5E2 --> x%2B7=-1%2F5 or x%2B7=1%2F5 ,
which can be summarized as x=-7+%2B-+1%2F5
So the horizontal axis would be 2%2F5 in length, the distance between those two extreme points. That is the major axis, the longest one.
The ends of the shorter vertical axis can be found for x=-7 where y=4+%2B-+1%2F8 ,
which makes the length of that axis 2%281%2F8%29=1%2F4 .

If the equation was %28x%2B7%29%5E2%2F5%5E2+%2B+%28y-4%29%5E2%2F8%5E2=+1 instead,
the ends of the horizontal axis would still be at y=4 ,
but with x=-7+%2B-+5 ,
and the ends of the vertical axis would still be at x=-7 ,
but with y=4+%2B-+8 .
The length of the axes would be
2%2A5=10 for the shorter, horizontal, minor axis,
and 2%2A8=16 for the longer, vertical, major axis.
graph%28300%2C300%2C-16%2C4%2C-6%2C14%2C%28x%2B7%29%5E2%2F25%2B%28y-4%29%5E2%2F64%3C=1%29