Question 1186658
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An ellipse has vertices (2 − √ 61, 5) and (2 + √ 61, 5), and its minor axis is 12 units long. 
Find its standard equation and its foci.
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            It can be done (and it should be done) in much shorter way,  than @MathLover1 does it.


            It does not require so intensive calculations.



<pre>
Looking at the foci coordinates, you see that they are on the horizontal line y = 5.


So, the major axis is horizontal, parallel to x-axis, and the length of the horizontal axis is  

     {{{2+sqrt(61)}}} - {{{(2-sqrt(61))}}} = {{{2*sqrt(61)}}}.


Hence, the length of the major semi-axis  "a"  is half of it, i.e.  a = {{{sqrt(61)}}}.


The length of the minor semi-axis is  b = {{{12/2}}} = 6.


The center of the ellipse is the point (2,5).


THEREFORE, the standard form equation of the ellipse is


    {{{(x-2)^2/61}}} + {{{(y-5)^2/6^2}}} = 1.    <U>ANSWER</U>


The distance from the center to the focus is  c = {{{sqrt(a^2 - b^2)}}} = {{{sqrt(61-36)}}} = {{{sqrt(25)}}} = 5.


The foci are  (2+5,5) = (7,5)  and  (2-5,5) = (-3,5).
</pre>

Solved.