Calc

A line tangent to y=(x^2)+1 at x=a, a>0, intersects the x-axis at point P.

a) Write an expression for the area of the triangle formed by the tangent line, the x-axis, and the line x=a.

b) For what value of 'a' is the area of the triangle a minimum? Justify your answer.

Show all work. Thanks!

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  1. quite a question for this forum.

    I worked it out, but will only give you some of the steps, leaving it up to you to fill in most of the steps.

    1. The graph is a parabola with vertex at (0,1)

    2. Let A(a,a^2 + 1) be the point of contact

    3. the slope of the tangent is the derivative of the function which is y' = 2x
    At the point A the slope is then 2a

    3. find the equation of the tangent line with slope 2a and point (a,a^2+1)

    let y = 0 and solve for x to get the x-intercept
    (you should get x = (a^2 - 1)/(2a)

    call that point P
    let the other point of the triangle which is on the x-axis be Q
    Of course Q is (a,0)

    So the base of the triangle is PQ and the height is AQ

    we need PQ which is a - (x-intercept)
    = a - (a^2 - 1)/(2a)

    you should get (a^2+1)/(2a), this is your base of the triangle
    the height AQ of course is a^2+1 from the equation.

    Now Area = (1/2)base x height

    Differentiate that, set it equal to zero and solve for a

    (I got a = 1/√3)

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    posted by Reiny

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