Question

How do you write an equation of an ellipse in standard form given vertices (4,-7) and (4,3) and foci (4,-6) and (4,2)?

Answer 1

`(x-3)^2/3^2-(y+2)^2/5^2=1` The Socratic graph is inserted.

Explanation:

graph{(x-3)^2/3^2+(y+2)^2/5^2-1=0 [-20, 20, -10, 10]}

The line joining the foci S(4, 2) and S'(4, -6) is the middle segment of

the the major axis joining the vertices A(4, 3) and A'(4, -7).

It follows that the common midpoint C(4, -2) is the center of the

ellipse, the major axis is along x = 4, in the (`yuarr`) and the minor

axis is along y = 2, in the `x rarr`. .

The distance between the vertices

`AA'= 2a = x_A-x_A'=3-(-7)=10`.,

giving a = 5. The distance between the foci

`SS'=2ae=10e=x_S-xS'=2-(-6)=8,` giving eccentricity

`e = 4/5`.

The semi minor-axis `= b =asqrt(1-e^2)=5sqrt(1-16/25)=3`.

Now, the equation of this ellipse with semi axes a = 5, b = 3, center

at C( 4, -2) and axes parallel to the y and x axes, respectively, is

`(x-3)^2/3^2-(y+2)^2/5^2=1`