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

Question

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!

Answer 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)