Question 147808
<font size = 9 color = "red">SOLUTION BY EDWIN</font></b>

 My question is: Write the equation for the ellipse whose center is the origin, has a horizontal major axis length of 12 and passses through the point (-4,2).
I need help with this, since the horizontal axis is at the origin I get that the foci is (-6,0) and (6,0). I find that the equation will be {{{(x^2/36)+(y^2/b^2)=1}}} but I am having trouble finding what {{{b^2}}} equals. Can you help me?

Go back and look at the problem to see what if anything you haven't used. Here's what you haven't used:

>>...passses through the point (-4,2)...<<

So you see that all you need do is plug the point (-4,2) into
{{{(x^2/36)+(y^2/b^2)=1}}} and solve for b.

{{{(x^2/36)+(y^2/b^2)=1}}}
{{{((-4)^2/36)+((2)^2/b^2)=1}}}

{{{16/36+4/b^2=1}}}

{{{4/9+4/b^2=1}}}

Clear of fractions by multiplying thru by {{{9b^2}}}

{{{9b^2*4/9+9b^2*4/b^2=9b^2*1}}}
 
{{{cross(9)b^2*4/cross(9)+9cross(b^2)*4/cross(b^2)=9b^2}}}

{{{4b^2+36=9b^2}}}

{{{36=5b^2}}}

{{{36/5=b^2}}}

That's an ugly answer, but it is correct.  So
the equation of the ellipse in standard form is

{{{x^2/36+y^2/(36/5)=1}}}


Here is its graph:

{{{drawing(400,400,-10,10,-10,10, grid(1), locate(-4-.2,2.4,"@"), graph(400,400,-10,10,-10,10,sqrt((36-x^2)/5)), locate(-6.4-.2,2.7,"(-4,2)"), 
graph(400,400,-10,10,-10,10,-sqrt((36-x^2)/5))
 
)}}}

But if you don't necessarily have to have it in standard
form, you can simplify it.

{{{x^2/36+y^2/(36/5)=1}}}

Multiply top and bottom of the second fraction by {{{5}}}

{{{x^2/36+5y^2/(5*(36/5))=1}}}

{{{x^2/36+5y^2/(cross(5)*(36/cross(5)))=1}}}

{{{x^2/36+5y^2/36=1}}}

Then multiply through by {{{36}}}

{{{x^2+5y^2=36}}}
 
Edwin</pre></font></b>