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# Advanced Math/Math - Homework Questions

Question

Math Question
I have a homework question for Math (Conics) that I can't seem to find an answer for...this is the only one I haven't been able to solve.

See attached image for the question I'm referring to.

Thank you very much for your assistance.
Sincerely yours,
S.

As I'm sure you can find in your textbook, the equation for a hyperbola offset from the origin by (x0,y0) and opening to the right and left is

(1)   (x-x0)^2/a^2 - (y-y0)^2/b^2 = 1.

The goal is to determine the values of the various parameters using the information given in the problem. To start, the offset y0 can easily seen to be 50 (from the coordinates for the point E and from the plot itself).

Perhaps the next best relationships to exploit are the equations for the asymptotes. As the name implies, these are the equations that describe the behavior when x and y approach infinity. Letting x' = x-x0 and y' = y-y0, the left side of eqn (1) overwhelms the right side to give

x'^2/a^2 - y^2/b^2 ≈ 0  or y = ±(b/a)x   ( the ± comes from taking the square root). This gives us the slope of the asymptotes in terms of a and b

(2)    b/a = 5/12  or  b = (5/12)a.

At this point, it is tempting to assume b = 5 and a = 12, but let's work thru it. Again, from the textbook, the foci of the hyperbola are located along the y = 50 axis at ±c, where

c^2 = a^2 + b^2 = a^2･(1 + 25/144)   using eq (2) to substitute for b.

The asymptotes cross at (x0, y0). The point E is one of the foci with the x-coordinate = 30, so we have

(3)   30 = x0 - c. From the geometry of the problem, one of the asymptotes goes from y0 = 50 at x = xo to 815/12 at x = 0, or

x0･(5/12) = ∆y = 815/12 - 50 = (815 - 600)/12 = 215/12  or  x0 = 215/5 = 43. From eq (3) we then have

c^2 = (13)^2 = 169, which gives us

a^2･(1 + 25/144) = 169  or  a = 12 and thus b = 5 (as we could have guessed per the above). Putting this all together, the eqn for the hyperbola is

(x - 43)^2/(12)^2 - (y-50)^2/(5)^2 = 1.

Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thank you! Very clear and thorough response that helped me understand how to solve these kinds of problems in the future. No complaints whatsoever and have a great day! :)

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#### randy patton

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