SOLUTION: find the center,transverse axis, vertices, foci and asymptotes of the hyperbola; 4x^2-49y^2=196

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the center,transverse axis, vertices, foci and asymptotes of the hyperbola; 4x^2-49y^2=196      Log On


   



Question 669841: find the center,transverse axis, vertices, foci and asymptotes of the hyperbola;

4x^2-49y^2=196

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
given: 4x%5E2-49y%5E2=196

First get it in standard form, which is either
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1 if the hyperbola opens right and left,
or
%28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1 if the hyperbola opens upward and downward.
4x%5E2%2F196-49y%5E2%2F196=1
x%5E2%2F49-y%5E2%2F4=1
x%5E2%2F7%5E2-y%5E2%2F2%5E2=1
note that:
the denominator of x%5E2 term is greater than that of the y term, so the major axis is horizontal

Since x%5E2=%28x-0%29%5E2 and y%5E2+=+%28y-0%29%5E2, then
h=0 and k=0
so, the center is at (h, k) = (0, 0).
a=7+and b=2,
the equation a%5E2+%2B+b%5E2+=+c%5E2 gives you c%5E2+=+49+%2B+4+=+53,
so c+=+sqrt%2853%29=+%287.28%29

and the eccentricity is e+=+7.28%2F7=1.04
the vertices and foci are above and below the center, so the foci are at
(0,-c) and (0,c), which is

(0,-7.28) and (0, 7.28)

and the vertices are at (0,a) and (0,-a) which is
(0,7) and (0,-7)

Because the x part of the equation is dominant , then the slope of the
asymptotes has the a on top, so the slopes will be m+=+%282x%2F7%29=%280.29%29.
so, the asymptotes are y=%280.29%29x and y=-%280.29%29x

graph:
graph%28+600%2C+600%2C+-15%2C+15%2C+-10%2C+10%2C+4x%5E2-49y%5E2%3C=196+%29%29