SOLUTION: Find the foci, eccentricity, length of latus rectum, and the x and y intercepts of the ellipse x^2/4 + y^2/16=1.

Question 1112231: Find the foci, eccentricity, length of latus rectum, and the x and y intercepts of the ellipse x^2/4 + y^2/16=1.
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
For a wider-than-tall ellipse with center at (,), having vertices units to either side of the center and foci units to either side of the center, the ellipse equation is:

For a taller-than-wide ellipse with center at (,), having vertices units above and below the center and foci units above and below the center, the ellipse equation is:
(y-h)^2/a^2 + (x-k)^2/b^2=1

The length of the semi-major axis is , a "semi-minor" axis is , the length of the whole major axis is ,the length of the whole minor axis is , and the distance between the foci is .
The three letters , , and are related by the equation .
you are given:
or

you also see that -> and ->
as you can see, and ; so, the center of your ellipse is at origin ()

now we can find the foci, eccentricity, length of latus rectum, and the x and y intercepts:
center is at: (, )
(, )
focus is the fixed value
(,) and (,)
use to find

or
so, foci is at:
(, ) and (, )
vertices at: (, ) and (, )
(, ) and (, )
covertices at: (, ) and (, )
(, ) and (, )
eccentricity is denoted as
which is

≈
length of latus rectum: =>
-intercept: set

or
-intercepts are at (, ) and (, )
-intercept: set

or
-intercepts are at (, ) and (, )

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