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Math Help - Hyperbola

  1. #1
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    Exclamation Hyperbola

    Find the acute angle between the tangents which can be drawn from (2,3) to the hyperbolas x^2 - y^2 = 2. I've found dy/dx = x/y but I don't know what to do with it. Do I simply sub in (2,3)? If so, what do I do next???
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Find the acute angle between the tangents which can be drawn from (2,3) to the hyperbolas x^2 - y^2 = 2. I've found dy/dx = x/y but I don't know what to do with it. Do I simply sub in (2,3)? If so, what do I do next???
    Well, you certainly don't substitute x= 2, y= 3 into the equation of the hyperbola because (2, 3) is NOT on the hyperbola! At any point, (x_0, y_0), the slope of the tangent line is given by m= \frac{dy}{dx}= \frac{x_0}{y_0} so the equation of a line that is tangent to the hyperbola at (x_0, y_0) is y= \frac{x_0}{y_0}(x- x_0)+ y_0.

    Now, since that tangent line must pass through (2, 3), you can replace x by 2 and y by 3 in that: 3= \frac{x_0}{y_0}(2- x_0)+ y_0. Multiplying through by y_0 to get rid of the fraction, 3y_0= 2x_0- x_0^2+ y_0^2 or x_0^2- y_0^2= 2x_0- 3y_0, a quadratic equation in x_0 and y_0. Since (x_0, y_0) is on the hyperbola, you also have x_0^2- y_0^2= 2 so that x_0^2- y_0^2= 2x_0- 3y_0= 2 and then y_0= (2/3)x_0- 2/3. You can substitute that into either of the previous equations to get a quadratic equation for x_0 which, hopefully, will have two roots giving the two points (x_0, y_0) where the two tangents through (2, 3) touch the hyperbola. That, of course, will give you the slopes, \frac{x_0}{y_0}, of the two lines and you can get the angle between them from that.
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  3. #3
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    Hmmmmm the coordinates I get are basically quadratic formula forms...how on earth do I solve that to get the angle??? is there any other possible "easier" way?
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