Question 284081
what is the equation for the conic section? it is an ellipse with the vertices
(-2,4) and (-2,-6). covertices are (-4,-1) and (0,-1). and what are the foci? 


<pre><font size = 4 color = "indigo"><b>
what is the equation for the conic section? it is an ellipse with the vertices
(-2,4) and (-2,-6). covertices are (-4,-1) and (0,-1). and what are the foci? 

<pre><font size = 4 color = "indigo"><b>
Plot those 4 points:
 
{{{drawing(400,400,-7,6,-7,6,
graph(400,400,-7,6,-7,6),
line(-2+.1,4,-2-.1,4), line(-2,4+.1,-2,4-.1),
line(-2+.1,-6,-2-.1,-6), line(-2,-6+.1,-2,-6-.1),
line(-4+.1,-1,-4-.1,-1), line(-4,-1+.1,-4,-1-.1),
line(0+.1,-1,0-.1,-1), line(0,-1+.1,0,-1-.1)
  )}}}
 
Connect them to show the major and minor axes
of the ellipse:
 
{{{drawing(400,400,-7,6,-7,6,
graph(400,400,-7,6,-7,6), line(-2,4,-2,-6), line(-4,-1,0,-1) )}}}
 
Sketch in the ellipse:
 
{{{drawing(400,400,-7,6,-7,6,
graph(400,400,-7,6,-7,6), line(-2,4,-2,-6), line(-4,-1,0,-1), 
arc(-2,-1,4,-10)


)}}}

Since the ellipse has its major axis vertical,
it has the standard form:
 
{{{(x-h)^2/b^2 + (y-k)^2/a^2 = 1}}}
 
where 
 
1. (h,k) = the center 
 
2. a = the distance from the center to either of the two vertices
 
3. b = the distance from the center to either of the the covertices.
 
We can see from the graph that 
 
1. the center of the ellipse is (h,k) = (-2,-1)
 
2. a = 5
 
3. b = 2
 
So the equation 
 
{{{(x-h)^2/b^2 + (y-k)^2/a^2 = 1}}}
 
becomes
 
{{{(x-(-2))^2/2^2 + (y-(-1))^2/5^2 = 1}}}
 
or
 
{{{(x+2)^2/4 + (y+1)^2/25 = 1}}}

to find the foci, we calculate c by
this Pythagorean relation:

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

{{{c^2=5^2-2^2}}}

{{{c^2=25-4}}}

{{{c^2=21}}}

{{{c=sqrt(21)}}}

The two foci are on the major axis c usits
from the center, so they are

(-2,-1+{{{sqrt(21)}}}) and (-2,-1-{{{sqrt(21)}}})

 
Edwin</pre>