9.5. Hyperbolas http://www.ck12.org
- Find the equation of the hyperbola that has foci at (13, 5) and (-17, 5) with asymptote slopes of±^34.
Answers: - 9x^2 − 16 y^2 − 18 x+ 96 y+ 9 = 0
9 (x^2 − 2 x)− 16 (y^2 − 6 y) =− 9
9 (x^2 − 2 x+ 1 )− 16 (y^2 − 6 y+ 9 ) =− 9 + 9 − 144
9 (x− 1 )^2 − 16 (y− 3 )^2 = 144
−(x−^1 )2
16 +(y− 3 )^2
9 =^1Shape: Hyperbola that opens vertically.
Center: (1, 3)
a= 3
b= 4
c= 5
e=ac=^53
d=ac^2 =^95
Foci: (1, 8), (1, -2)
Vertices: (1, 6), (1, 0)
Equations of asymptotes:(x− 1 ) =±^34 (y− 3 )
Note that it is easiest to write the equations of the asymptotes in point-slope form using the center and the slope.
Equations of directrices: y= 3 ±^95
- Since exact points are not marked, you will need to estimate the slope of asymptotes to get an approximation fora
andb. The slope seems to be about±^23. The center seems to be at (-1, -2). The transverse axis is 6 which means
a=3.
(x+ 91 )^2 −(y+ 42 )^2 = 1 - The center of the conic must be at (-2, 5). The focal radius isc=15. The slopes of the asymptotes are±^34 =ba.
a^2 +b^2 =c^2