Question 145137
Find the equation of an ellipse with its center at (1,2), focus at (6, 2) and containing the point (4,6). 
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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:

{{{(x-h)^2/a^2+(y-k)^2/b^2=1}}}

The center is ({{{h}}},{{{k}}})

The left focus is ({{{h-c}}},{{{k}}}) and the 
right focus is ({{{h+c}}},{{{k}}}), where {{{c^2=a^2-b^2}}}

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

{{{(x-1)^2/a^2+(y-2)^2/b^2=1}}}

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

{{{(4-1)^2/a^2+(6-2)^2/b^2=1}}}

{{{3^2/a^2+4^2/b^2=1}}}

{{{9/a^2+16/b^2=1}}}

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

({{{h+c}}},{{{k}}}) = ({{{6}}},{{{2}}}) 
  
So {{{h+c = 6}}} and of course {{{k=2}}}and since {{{h = 1}}}

   {{{1+c = 6}}}

     {{{c = 5}}}

Now using the fact that {{{c^2=a^2-b^2}}} substituting {{{c=5}}}:

{{{5^2=a^2-b^2}}}
{{{25=a^2-b^2}}}
{{{a^2-b^2=25}}}

So we have this system of equations:


{{{9/a^2+16/b^2=1}}}
{{{a^2-b^2=25}}}

Can you solve that for {{{a^2}}} and {{{b^2}}} by substitution? 
If not post again asking how.

You get {{{a^2=45}}} and {{{b^2=20}}}

Substituting in 

{{{(x-1)^2/a^2+(y-2)^2/b^2=1}}}

{{{(x-1)^2/45+(y-2)^2/20=1}}}

Edwin</pre>