SOLUTION: Find the standard form of the equation of each ellipse satisfying the given conditions; Endpoints of major axis:(7,9) and (7,3) Endpoints of minor axis: (5,6) and (9

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Question 191554This question is from textbook Blitzer College Algebra an early funcitons approach
: Find the standard form of the equation of each ellipse satisfying the given conditions; Endpoints of major axis:(7,9) and (7,3)
Endpoints of minor axis: (5,6) and (9,6)
This question is from textbook Blitzer College Algebra an early funcitons approach

Answer by Edwin McCravy(6941) About Me  (Show Source):
You can put this solution on YOUR website!

Plot those 4 points:

drawing%28400%2C400%2C-2%2C11%2C-2%2C11%2C%0D%0Agraph%28400%2C400%2C-2%2C11%2C-2%2C11%29%2C%0D%0Aline%284.9%2C6%2C5.1%2C6%29%2C+line%285%2C5.9%2C5%2C6.1%29%2C%0D%0Aline%288.9%2C6%2C9.1%2C6%29%2C+line%289%2C5.9%2C9%2C6.1%29%2C%0D%0Aline%286.9%2C3%2C7.1%2C3%29%2C+line%287%2C2.9%2C7%2C3.1%29%2C%0D%0Aline%286.9%2C9%2C7.1%2C9%29%2C+line%287%2C8.9%2C7%2C9.1%29%0D%0A++%29

Connect them to show the major and minor axes
of the ellipse:

drawing%28400%2C400%2C-2%2C11%2C-2%2C11%2C%0D%0Agraph%28400%2C400%2C-2%2C11%2C-2%2C11%29%2C+line%285%2C6%2C9%2C6%29%2C+line%287%2C3%2C7%2C9%29+%29

Sketch in the ellipse:

drawing%28400%2C400%2C-2%2C11%2C-2%2C11%2C%0D%0Agraph%28400%2C400%2C-2%2C11%2C-2%2C11%29%2C+line%285%2C6%2C9%2C6%29%2C+line%287%2C3%2C7%2C9%29%2C+%0D%0Agraph%28400%2C400%2C-2%2C11%2C-2%2C11%2C6%2B%283%2F2%29sqrt%284-%28x-7%29%5E2%29%29%2C%0D%0Agraph%28400%2C400%2C-2%2C11%2C-2%2C11%2C6-%283%2F2%29sqrt%284-%28x-7%29%5E2%29%29%0D%0A%29

We can see that the ellipse has the standard form:

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

where 

1. (h,k) = the center 

2. a = the distance from the center to either end of the 
major axis.

3. b = the distance from the center to either end of the 
minor axis.

We can see from the graph that 

1. the center of the ellipse is (h,k) = (7,6)

2. a = 3

3. b = 2

So the equation 

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

becomes

%28x-7%29%5E2%2F2%5E2+%2B+%28y-6%29%5E2%2F3%5E2+=+1

or

%28x-7%29%5E2%2F4+%2B+%28y-6%29%5E2%2F9+=+1

Edwin