SOLUTION: Name the center,foci,and the vertics of each ellipse whose equation is given (x-3)^2/25 + (y-4)^2/16 =1

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Question 83900This question is from textbook Algebra and Trigonometry
: Name the center,foci,and the vertics of each ellipse whose equation is given
(x-3)^2/25 + (y-4)^2/16 =1 This question is from textbook Algebra and Trigonometry

Answer by chitra(359) About Me  (Show Source):
You can put this solution on YOUR website!
The given equation is: %28%28x-3%29%5E2%2F25%29+%2B+%28%28y-4%29%5E2%2F16%29+=+1

Comparing this to the standard equation, we get:

%28%28x+-+h%29%5E2%2Fa%5E2%29+%2B+%28%28y+-+k%29%5E2%2Fb%5E2%29+=+1

We find that the centre of the ellipse given by (h,k) is equal to (3, 4)

The foci of the Ellipse is given by (+- ae, 0)

So we first find the eccentricity using the formula:

e = +sqrt%281+-+%28b%5E2%2Fa%5E2%29%29+

e = +sqrt%281+-+16%2F25%29%29+

e = 3/5

So now focus is: (+- ae, 0) = (+ -3, 0)

Now the vertex is (+ - a, 0) = (+- 5, 0)

hence, the solution