Asked by Anonymous
                The slope of the tangent line to a curve at any point (x, y) on the curve is x divided by y. What is the equation of the curve if (3, 1) is a point on the curve?
            
            
        Answers
                    Answered by
            Reiny
            
    slope = x/y = 3/1 = 3
so the equation of the tangent at the point (3,1) is
y-1 = 3(x-3)
y = 3x -8 or 3x - y - 8 = 0
    
so the equation of the tangent at the point (3,1) is
y-1 = 3(x-3)
y = 3x -8 or 3x - y - 8 = 0
                    Answered by
            Steve
            
    y' = x/y
you can recognize that as an hyperbola
x^2-y^2 = a^2
since (3,1) is on the curve,
9-1 = a^2
x^2-y^2 = 8
    
you can recognize that as an hyperbola
x^2-y^2 = a^2
since (3,1) is on the curve,
9-1 = a^2
x^2-y^2 = 8
                    Answered by
            Reiny
            
    Maybe I should become a politician.
Never actually answer the question, rather give an answer to some other question.
Thanks for the catch Steve, that is two for one evening, I should get more sleep.
    
Never actually answer the question, rather give an answer to some other question.
Thanks for the catch Steve, that is two for one evening, I should get more sleep.
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