Find the equation of the hyperbola with a conjugate parallel to the y-axis given the center at (2,-5), slopes of the asymptote numerically one-sixteenth times the length of the latus rectum, and the distance between the foci is 2sqrt145

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The graph looks complicated but it contains enough information to answer the question.

The origin of the graph is marked (h,k) representing a displacement from (0,0). All the text is based on hyperbolas that have their origin at (0,0) to keep the formulas as simple as possible.

Colour-coding aids legibility: the green arrows and text show the latus recta (LR1 to LR4); the red arrows and text show the directrices (D1 to D4); blue shows foci, a and b in the hyperbola formulae; shades of grey show the hyperbolas, axes, asymptotes and grid lines. The yellow crosses mark the foci F1 to F4.

The left-right orientated hyperbola is the “normal” hyperbola with its transverse (major) axis as the x-axis. The associated conjugate hyperbola is up-down and the conjugate (minor) axis is the y-axis. Both hyperbolas have the same asymptotes shown as y=bx/a and y=-bx/a. The centre is at (h,k)=(2,-5). One hyperbola is the conjugate of the other. F1, F2, LR1, LR2, D1, D2 apply to the hyperbola (normal left-right), while the remaining labels apply to the conjugate hyperbola.

The distance between the foci is 2√a²+b² as can be seen by the yellow crosses. So √a²+b²=√145 and a²+b²=145. The numerical value of the asymptote slope is b/a. “Conjugate parallel to the y-axis” I understand to mean the conjugate axis is parallel to the y-axis, so the transverse axis of the hyperbola is parallel to the x-axis. The equation of the conjugate axis is x=h=2, and that of the transverse axis is y=k=-5. This means we need the equation of the left-right hyperbola for which LR=2b²/a. So b/a=2b²/16a=b²/8a, and b=8. a²+b²=145, a²=145-64=81.

We need to adjust the origin, or centre, of the hyperbola so in place of x we need x-5 and in place of y, y+5. 

The equation of the hyperbola is (x-2)²/81-(y+5)²/64=1.

 

by Top Rated User (1.1m points)

Find the equation of the hyperbola with a conjugate parallel to the y-axis given the center at (2,-5), slopes of the asymptote numerically one-sixteenth times the length of the latus rectum, and the distance between the foci is 2sqrt145

Latus rectum is 2p

And p = b^2/a  ------------------------------------- (1)

Distance between foci is 2c = 2√(145)

And b^2 = c^2 – a^2   ---------------------------- (2)

Slope of asymptote is |m| = b/a

And m = (1/16)*(2p)   (using positive slope)

b/a = p/8    ------------------------------------------ (3)   

Using (3) and (1)

8b/a = p = b^2/a

b = 8

Using (2) with c = √(145),

64 = 145 – a^2

a^2 = 81

a = 9

Eqn of Hyperbola

x^2/a^2 – y^2/b^2 = 1

x^2/81 – y^2/64 = 1        (Centred at (0,0))

(x – 2)^2/81 – (y + 5)^2/64 = 1         (Centred at (2, -5))

by Level 11 User (81.5k points)

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