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)?

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Answer 1 · 2017-01-09T14:26:21

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

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

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