Unformatted text preview: righthand side equal to 1.
9y 2 − x2 − 6x
9y 2 − 1 x2 + 6x
9y 2 − x2 + 6x + 9
9y 2 − (x + 3)2
y 2 (x + 3)2
1−
1
9 =
=
=
= 10
10
10 − 1(9)
1 =1 Now that this equation is in the standard form of Equation 7.7, we see that x − h is x + 3 so
h = −3, and y − k is y so k = 0. Hence, our hyperbola is centered at (−3, 0). We ﬁnd that a2 = 1
so a = 1, and b2 = 1 so b = 1 . This means that we move 1 unit to the left and right of the
9
3
center and 1/3 units up and down from the center to arrive at points on the guide rectangle. Since
the x2 term is being subtracted from the y 2 term, we know the branches of the hyperbola open
upwards and downwards. This means the transverse axis lies along the vertical line x = −3 and the
conjugate axis lies along the xaxis. Since the vertices of the hyperbola are where the hyperbola
intersects the transverse axis, we get that the vertices are 1 of a unit above and below (−3, 0) at
3 440 Hooked on Conics 1
−3, 1 and −3, − 3 . To ﬁnd the foci, we need c =
3 √ 1
9 a2 + b2 = √ lie on the transverse axis, we move
√ −3, − 10
3 10
3 √ +1 = 10
3. S...
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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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