Question 1187462
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foci (-2, 3) and (4, 3) b=4
Please help me solve this. Find the general and standard form then graph it. 
If it is not that much of a hassle, i would also appreciate it so much if it has a solution. 
This is all about ellipse. This is the equation that i ended up with  {{{(x-1)^2/25+(y-3)^2/16=1}}}
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The method of analysis is close to what I showed you for similar problem under this link


<A HREF=https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Quadratic-relations-and-conic-sections.faq.question.1187463.html>https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Quadratic-relations-and-conic-sections.faq.question.1187463.html</A>


https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Quadratic-relations-and-conic-sections.faq.question.1187463.html




<pre>
The foci lie on the horizontal line y= 3 --- hence, the major axis of the ellipse is HORIZONTAL.


The center of the ellipse is the point  (1,3) ---  it explains the numbers in the equation of the ellipse.


Further, the distance between the foci is  4 - (-2) = 6 units;  hence, the eccentricity of the ellipse is  c = {{{6/2}}} = 3 units.


Now you know "c" and "b" of the ellipse:  c = 3, b = 4 (given).  (As a reminder: b is the minor semi-axis).


You are in position to find the major semi-axis  a = {{{sqrt(b^2 + c^2)}}} = {{{sqrt(4^2+3^2)}}} = {{{sqrt(25)}}} = 5.


After that, the equation of the ellipse is


    {{{(x-1)^2/25}}} + {{{(y-3)^2/16}}} = {{{1}}}.
</pre>

Solved, and everything is explained.


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For basic info about ellipses, &nbsp;see my 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.