SOLUTION: Find the equation of an ellipse with its center at (1,2), focus at (6, 2) and containing the point (4,6). I have so far (x-1)^2 + (y-2)^2 = 1, but I cannot figure out what my

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the equation of an ellipse with its center at (1,2), focus at (6, 2) and containing the point (4,6). I have so far (x-1)^2 + (y-2)^2 = 1, but I cannot figure out what my       Log On


   



Question 145137: Find the equation of an ellipse with its center at (1,2), focus at (6, 2) and containing the point (4,6).
I have so far (x-1)^2 + (y-2)^2 = 1, but I cannot figure out what my a^2 and b^2 should be for my denominators. Please help.

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation of an ellipse with its center at (1,2), focus at (6, 2) and containing the point (4,6).

Since the center and focus have the same y-value, 
we know this is an ellipse whose major axis is 
horizontal, so its equation is in the form:

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

The center is (h,k)

The left focus is (h-c,k) and the 
right focus is (h%2Bc,k), where c%5E2=a%5E2-b%5E2

---------------------------------------------

We substitute the center (h,k) = (1,2)

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

We substitute the given point (x,y) = (4,6)

%284-1%29%5E2%2Fa%5E2%2B%286-2%29%5E2%2Fb%5E2=1

3%5E2%2Fa%5E2%2B4%5E2%2Fb%5E2=1

9%2Fa%5E2%2B16%2Fb%5E2=1

The focus given (6,2) must be a right focus
since (6,2) is right of the center (1,2), so

(h%2Bc,k) = (6,2) 
  
So h%2Bc+=+6 and of course k=2and since h+=+1

   1%2Bc+=+6

     c+=+5

Now using the fact that c%5E2=a%5E2-b%5E2 substituting c=5:

5%5E2=a%5E2-b%5E2
25=a%5E2-b%5E2
a%5E2-b%5E2=25

So we have this system of equations:


9%2Fa%5E2%2B16%2Fb%5E2=1
a%5E2-b%5E2=25

Can you solve that for a%5E2 and b%5E2 by substitution? 
If not post again asking how.

You get a%5E2=45 and b%5E2=20

Substituting in 

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

%28x-1%29%5E2%2F45%2B%28y-2%29%5E2%2F20=1

Edwin