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Find the equation of the tangent line and the normal line to the graph of the equation at the indicated point.

x^2-y^2=16, (5,3)

I need to show work, so answers formatted in this manner would be most appreciated. Thanks! :) :)

  • Calculus -

    For the tangent line slope, you want dy/dx.
    In your case you can differentiate both sides of the equation with respect to x, treating y as a function of x.
    2x - 2 dy*dy/dx = 0
    dy/dx = x/y = 5/3

    For the equation of the line, write it in the form
    (y -3)/(x-5) = slope = 5/3
    y-3 = (5/3)(x-5) = 5/3 x -25/3
    y = (5/3)x -16/3

  • Calculus -

    Thanks so much! :)

    You're a lifesaver!

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