SOLUTION: write the equation. 1. hyperbola with foci (0,4) and (0,-4) and asymptotes at y=2x 2. circle passing through the points (12,1) and (2,-3) with center on the line 2x-5y+10=0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: write the equation. 1. hyperbola with foci (0,4) and (0,-4) and asymptotes at y=2x 2. circle passing through the points (12,1) and (2,-3) with center on the line 2x-5y+10=0      Log On


   



Question 1035355: write the equation.
1. hyperbola with foci (0,4) and (0,-4) and asymptotes at y=2x
2. circle passing through the points (12,1) and (2,-3) with center on the line 2x-5y+10=0

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1.The major axis is on the line connecting the foci, which is the y-axis.
The center of the hyperbola is halfway between the foci, at (0,0), the origin,
so the minor axis (perpendicular to the y-axis through the center) is the x-axis.
The equation for a parabola centered at the origin, with the y-axis as its major axis is
y%5E2%2Fa%5E2-x%5E2%2Fb%5E2=1 with a%3E0 and b%3E0 .
The focal distance, c , is half the distance between the foci.
In this case, it is c=%284-%28-4%29%29%2F2=4 .
The semimajor axis, a , is the distance between the center and each vertex.
The major axis is the vertical segment connecting vertices (0,-a), (0,a),
which are on the major axis, between the foci, so 0%3Ca%3C4 .
The semi-minor axis is b .
The asymptotes cross at the center of the hyperbola (in this case (0,0), the origin),
and have equations y=%28a%2Fb%29x and -%28a%2Fb%29x .
So far, we have

The values a and b determine a rectangle,
passing through the vertices,
with sides measuring 2a and 2b ,
and diagonals measuring 2c .
The axes and the asymptotes divide that rectangle into 8 right trianlges with legs a and b , and hypotenuse c , so
c%5E2=a%5E2%2Bb%5E2 .
Since the asymptotes have equations y=2x and y=-2x ,
in this case a%2Fb=2 --> a=2b --> a%5E2=4b%5E2 ,
and since c=4 ,
4%5E2=4b%5E2%2Bb%5E2
16=5b%5E2
b%5E2=16%2F5
a%5E2=4%2A16%2F5=64%2F5 .
So, the equation for the hyperbola in this problem is
y%5E2%2F%2864%2F5%29-x%5E2%2F%2816%2F5%29=1 ,
or y%5E2%2F12.8-x%5E2%2F3.2=1 ,
or 5y%5E2%2F64-5x%5E2%2F16=1 ,
or y%5E2%2F64-c%5E2%2F16=1%2F5 .