SOLUTION: locate the center, foci, vertices, and ends of the latera recta of the ellipse. find the equation of the ellipse satisfying the given conditions. find the equation for the specifi

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: locate the center, foci, vertices, and ends of the latera recta of the ellipse. find the equation of the ellipse satisfying the given conditions. find the equation for the specifi      Log On


   

Question 605623: locate the center, foci, vertices, and ends of the latera recta of the ellipse. find the equation of the ellipse satisfying the given conditions.
find the equation for the specified hyperbola center at the origin, latus rectum 64/3, eccentricity 5/3. pls graph it
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
EQUATION FOR THE HYPERBOLA:
A hyperbola with the equation x%5E2%2Fa%5E2-y%5E2%2Fb%5E2=1
has an eccentricity of
e=sqrt%281%2Bb%5E2%2Fa%5E2%29
and a latus rectum of 2%2Al=2b%5E2%2Fa
So from 2b%5E2%2Fa=64%2F3 we get b%5E2%2Fa=32%2F3
We also have
sqrt%281%2Bb%5E2%2Fa%5E2%29=5%2F3 --> 1%2Bb%5E2%2Fa%5E2=25%2F9 --> b%5E2%2Fa%5E2=25%2F9-1 --> b%5E2%2Fa%5E2=16%2F9 --> b%2Fa=4%2F3
Combining both equations:
%28b%5E2%2Fa%29%2F%28b%2Fa%29=%2832%2F3%29%2F%284%2F3%29 --> highlight%28b=8%29
and then
b%2F%28b%2Fa%29=8%2F%284%2F3%29=8%2A3%2F4 --> highlight%28a=6%29
That gives us the equation:
x%5E2%2F6%5E2-y%5E2%2F8%5E2=1 or highlight%28x%5E2%2F36-y%5E2%2F64=1%29

GRAPHING:
With a=6 and b=8, we can draw that box that gives us the vertices and the asymptotes.

We also know that
e=c%2Fa so 5%2F3=c%2F6 --> highlight%28c=10%29
which tells us that the foci are at distance 10 from the center.
And since the latus rectum is 64/3, there are points of the hyperbola, 32/3 above and below the foci.
That gives us the location of the foci and 4 more points of the hyperbola
I have the foci (green circles) and six points of the hyperbola (blue circles marking the vertices and the ends of the latera recta).
Now, I would just connect the points of the hyperbola that I found with smooth curved arches ) ( that hug the asymptotes towards their ends.
graph%28300%2C300%2C-25%2C25%2C-25%2C25%2C4x%2F3%2C-4x%2F3%2Cx%5E2%2F36-y%5E2%2F64%3E1%29