SOLUTION: Find the coordinate of the vertices and foci and the equation of the asymptotes for the hyperbola with the equation (x^2/81)-(y^2/49)=1

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Question 1142075: Find the coordinate of the vertices and foci and
the equation of the asymptotes for the hyperbola
with the equation (x^2/81)-(y^2/49)=1
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


x%5E2%2F81-y%5E2%2F49+=+1

%28x-0%29%5E2%2F9%5E2-%28y-0%29%5E2%2F7%5E2+=+1

%28x-0%29%5E2%2Fa%5E2-%28y-0%29%5E2%2Fb%5E2+=+1

The center of the hyperbola is (0,0).

The x squared term is positive, so the branches open right and left.

The length of the transverse axis (in the x direction, between the two vertices) is 2*a = 2*9=18; the length of the conjugate axis (the perpendicular bisector of the transverse axis) is 2*b = 2*7=14.

To sketch the hyperbola, draw a rectangle with the ends of the transverse and conjugate axes as the midpoints of the sides of the rectangle. The diagonals of that rectangle are the asymptotes of the hyperbola.

a is the distance from the center to each vertex.

c is the distance from the center to each focus, where c^2 = a^2+b^2.

That should give you all the information you need to answer the questions.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
For the hyperbola given by equation


    x%5E2%2F81 - y%5E2%2F49 = 1,     (1)


the equation for the asymptotes is


    x%5E2%2F81 - y%5E2%2F49 = 0.      (2)


It is obtained from the standard equation of the hyperbola (1) replacing "1" in the right side of the hyperbola' standard equation by "0".


The equation (2) deploys in two equations (3) and (4) below for the two straight asymptotes in this way


    %28x%2F9+%2B+y%2F7%29%2A%28x%2F9+-+y%2F7%29 = 0  =================>


    x%2F9+%2B+y%2F7 = 0    (3),   and

    x%2F9+-+y%2F7 = 0.   (4)


You can further transform these equations for asymptotes (3), (4) to any other appropriate equivalent form.