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Homework Help: Find an Equation of the Tangent Lines Passing Through a Poin

  1. Aug 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Find an equation of the tangent lines [in point-slope form] to the hyperbola [itex]x^2 + y^2 = 16[/itex] that pass through the point (2, -2).

    2. Relevant equations
    Point-slope form: [itex]y - y_1 = m(x - x_1)[/itex]
    Slope: [itex]m = \frac {y_1 - y_0} {x_1 - x_0}[/itex]

    3. The attempt at a solution
    First, take the derivative using implicit differentiation.

    [itex]D_x: 2x + 2yy' = 0 \Rightarrow y' = \frac {x} {y}[/itex]

    Since the derivative is the general form of the slope of a line at any of its points, then we can put it equal to the equation for slope. Furthermore, the derivative is tangent to a particular point on the original equation, which we can call point [itex](x_1, y_1)[/itex]. Since we found the equation of the slope at this point to be [itex]y' = \frac {x} {y}[/itex], then plugging in the point [itex](x_1, y_1)[/itex] (a particular case) should also work: [itex]y' = \frac {x_1} {y_1}[/itex]. Finally, we were given our second point above as (2, -2).

    [itex]y' = \frac {x_1} {y_1} = \frac {y_1 + 2} {x_1 - 2} \Rightarrow x_1^2 - 2x_1 = y_1^2 + 2y_1 \Rightarrow 2x_1 - 2y_1 = x_1^2 + y_1^2[/itex]

    Since the general equation [itex]x^2 + y^2 = 16[/itex] works for any pair of points on the line, then it should also work for the particular case of the point on the tangent line that is intersecting it, [itex](x_1, y_1)[/itex].

    [itex]2x_1 - 2y_1 = x_1^2 + y_1^2 = 16 \Rightarrow x_1 - y_1 = 8 \Rightarrow y_1 = x_1 - 8[/itex]

    Substitute into [itex]x_1^2 + y_1^2 = 16[/itex] and solve for [itex]x_1[/itex] and [itex]y_1[/itex]. We find [itex]x_1 = 5[/itex], [itex]y_1 = 3[/itex] and [itex]m = \frac 5 3[/itex]. Hence, the equation of the tangent line is [itex]y - 3 = \frac 5 3 (x - 5)[/itex].

    It took me awhile to reason through this problem, so I wasn't sure if it was correct. Thank-you.
  2. jcsd
  3. Aug 16, 2015 #2


    Staff: Mentor

    What you wrote is a circle. Did you mean ##x^2 - y^2 = 16##?
    The above isn't consistent with the equation you showed (which itself could be incorrect). If you differentiate ##x^2 + y^2 = 16## w.r.t. x, you get 2x + 2yy' = 0, or y' = -x/y.
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