Question 1204640
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Write the equation for the ellipse in standard form and general form.
foci at (-1,-1) and (9,-1), sum of focal radii 26
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<pre>
From given information about the foci coordinates, you can see that the 
distance between the foci points is 9 - (-1) = 10.


So, the half of this distance is 10/2 = 5 units.


This distance is the eccentricity, which has the standard designation "c".

So, c = 5 units.



Next, since the sum of focal radii is 26, it means that the distance from 
any focus of the ellipse to its any co-vertex is 26/2 = 13.


Thus we have a right angled triangle with one leg of 5 units and the hypotenuse of 13 units.

So, the other leg is  {{{sqrt(13^2-5^2)}}} = {{{sqrt(169-25)}}} = {{{sqrt(144)}}} = 12 units.

Thus we found the minor semiaxis of the ellipse: it is 12 units.

The standard designation for the minor semi-axis of an ellipse is "b".

So, for our ellipse b = 12 units.


Finally, if "a" is the major semi-axis, then we have

    {{{c^2}}} = {{{a^2}}} - {{{b^2}}},

or

    {{{5^2}}} = {{{a^2}}} - {{{12^2}}},

    {{{a^2}}} = 25 + 144 = 169

which implies

    a = {{{sqrt(169)}}} = 13.


Thus the major semi-axis is 13 units long.


Now the standard form of this ellipse equation is

    {{{(x-4)^2/13^2}}} + {{{(y+1)^2/12^2}}} = 1.


It is because the center of the ellipse is at the point (4,-1).
</pre>

Solved.


After that, to find the general equation is simple arithmetic.


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For basic info about ellipses, see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Ellipse-definition--canonical-equation--characteristic-points-and-elements.lesson>Ellipse definition, canonical equation, characteristic points and elements</A>  

in this site.