Notice that on the first page this problem reads this way:

Quadratic-relations-and-conic-sections/1169036 (2020-11-05 01:23:06): 

See the " 2020-11-05 01:23:06 "?  This means that this problem was posted by a 
student 4 1/2 years ago on November 5, 2020 at 1:23:06 AM.  The student who
posted this is no doubt long gone from the class he or she was taking back in
2020, and is no longer a student. 

Back in 2020 there were still lots of algebra and trig problems being posted.
There weren't enough tutors on here, and many problems like this one scrolled
off before the tutors got around to solving them because there were so many.
    
But then the schools learned that kids can take basic statistics with almost no
knowledge of second year algebra, no geometry, or trig. First year algebra
is enough mathematics for basic statistics, especially now since it's mostly all
done on computers.    

So apparently, now most high school students take statistics instead of trig and
college algebra.  That's why practically all posts on here nowadays are
statistics problems.  

OK, I'll shut up and do this problem posted back in 2020.  The former students
getting solutions in their emails to problems they posted back in 2020 are
probably having a good laugh!  J

----------------------

Find the equation of the ellipse if the endpoints of minor axis are (1,3) and
(1,-1) with a focus at (-1,1).

First draw what is given, the minor axis, and plot a focus at (-1,1).



Since the minor axis is vertical, we know that the major axis is horizontal,
and the ellipse is like the cross section of an egg sitting on a table, [and
not like the number 0, when the minor axis is horizontal and the major axis is
vertical.] So the length of the semi-major axis ' a ' goes under the term with
x, and the length of the semi-minor axis ' b ' goes under the term with y.  So
the form of the equation is

 

Where the center is the point (h,k).  

Next, we draw the center, which is the midpoint of the minor axis. We can tell
the center is the point (1,1).




so we know that the center (h,k) = (1,1).

So we have everything but ' a '. So we can write this much of the equation:



No we calculate ' a '.

The value " c " is the distance from the center to a focus. We can tell that the
distance from the center (0,0) to a focus (-1,1) is 2 units, so the value of 
c = 2.  Then we use the Pythagorean fact of all ellipses, which is

 where 

a is the length of the semi-major axis,
b is the length of the semi-minor axis, and
c is the distance from the center to a focus.

Since the minor axis that we drew is 4 units long, the semi-minor axis, b, is
half that, or b=2.  So we substitute b=2 and c=2 into






So now we know the equation is



or

      <----ANSWER!

But let's finish drawing the ellipse, because you will be asked to do that:

 which is about 2.8.

So draw the major axis, which goes from about 2.8 units left of the center,
to about 2.8 units right of the center. 




Now we can sketch in the ellipse. 




And we might as well plot the other focus which is 2 units on the right of the 
center.



Edwin