Here's the problem:
Let A be any positive number. Let l[sub:1gw6ypfr]1[/sub:1gw6ypfr] be the line through the origin of slope A and let l[sub:1gw6ypfr]2[/sub:1gw6ypfr] be the tangent line to the curve y = 1/x (x > 0) of slope – A. Show that the area of the triangle cut off by l[sub:1gw6ypfr]1[/sub:1gw6ypfr], l[sub:1gw6ypfr]2[/sub:1gw6ypfr], and the x-axis is equal to 1.
My work so far:
The equation of l[sub:1gw6ypfr]1[/sub:1gw6ypfr] is y = Ax. If we let (x[sub:1gw6ypfr]2[/sub:1gw6ypfr], y[sub:1gw6ypfr]2[/sub:1gw6ypfr]) be the point on l[sub:1gw6ypfr]2[/sub:1gw6ypfr] where it touches the curve y = 1/x (x > 0), the equation of l[sub:1gw6ypfr]2[/sub:1gw6ypfr] is y - 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr] = - A (x - x[sub:1gw6ypfr]2[/sub:1gw6ypfr]). Let (x[sub:1gw6ypfr]1[/sub:1gw6ypfr], y[sub:1gw6ypfr]1[/sub:1gw6ypfr]) be the point where l[sub:1gw6ypfr]1[/sub:1gw6ypfr] and l[sub:1gw6ypfr]2[/sub:1gw6ypfr] intersect, that is, where Ax = - Ax + Ax[sub:1gw6ypfr]2[/sub:1gw6ypfr] + 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr], or where x[sub:1gw6ypfr]1[/sub:1gw6ypfr] = (Ax[sub:1gw6ypfr]2[/sub:1gw6ypfr] + 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr])/2A. The height of the triangle is the value of y[sub:1gw6ypfr]1[/sub:1gw6ypfr] = (Ax[sub:1gw6ypfr]2[/sub:1gw6ypfr] + 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr])/2. The base of the triangle is the value x where l[sub:1gw6ypfr]2[/sub:1gw6ypfr] crosses the x-axis, that is, where - Ax + Ax[sub:1gw6ypfr]2[/sub:1gw6ypfr] + 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr] = 0. This occurs when x = (Ax[sub:1gw6ypfr]2[/sub:1gw6ypfr] + 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr])/A.
At this point, it's not obvious to me how (base * height)/2 = 1 from the calculations I've done. This problem is an exercise in a section of my textbook on derivatives of products and quotients. Since I haven't used any derivatives of products or quotients in my solution to this exercise yet, I think I must be on the wrong track.
Let A be any positive number. Let l[sub:1gw6ypfr]1[/sub:1gw6ypfr] be the line through the origin of slope A and let l[sub:1gw6ypfr]2[/sub:1gw6ypfr] be the tangent line to the curve y = 1/x (x > 0) of slope – A. Show that the area of the triangle cut off by l[sub:1gw6ypfr]1[/sub:1gw6ypfr], l[sub:1gw6ypfr]2[/sub:1gw6ypfr], and the x-axis is equal to 1.
My work so far:
The equation of l[sub:1gw6ypfr]1[/sub:1gw6ypfr] is y = Ax. If we let (x[sub:1gw6ypfr]2[/sub:1gw6ypfr], y[sub:1gw6ypfr]2[/sub:1gw6ypfr]) be the point on l[sub:1gw6ypfr]2[/sub:1gw6ypfr] where it touches the curve y = 1/x (x > 0), the equation of l[sub:1gw6ypfr]2[/sub:1gw6ypfr] is y - 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr] = - A (x - x[sub:1gw6ypfr]2[/sub:1gw6ypfr]). Let (x[sub:1gw6ypfr]1[/sub:1gw6ypfr], y[sub:1gw6ypfr]1[/sub:1gw6ypfr]) be the point where l[sub:1gw6ypfr]1[/sub:1gw6ypfr] and l[sub:1gw6ypfr]2[/sub:1gw6ypfr] intersect, that is, where Ax = - Ax + Ax[sub:1gw6ypfr]2[/sub:1gw6ypfr] + 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr], or where x[sub:1gw6ypfr]1[/sub:1gw6ypfr] = (Ax[sub:1gw6ypfr]2[/sub:1gw6ypfr] + 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr])/2A. The height of the triangle is the value of y[sub:1gw6ypfr]1[/sub:1gw6ypfr] = (Ax[sub:1gw6ypfr]2[/sub:1gw6ypfr] + 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr])/2. The base of the triangle is the value x where l[sub:1gw6ypfr]2[/sub:1gw6ypfr] crosses the x-axis, that is, where - Ax + Ax[sub:1gw6ypfr]2[/sub:1gw6ypfr] + 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr] = 0. This occurs when x = (Ax[sub:1gw6ypfr]2[/sub:1gw6ypfr] + 1/x[sub:1gw6ypfr]2[/sub:1gw6ypfr])/A.
At this point, it's not obvious to me how (base * height)/2 = 1 from the calculations I've done. This problem is an exercise in a section of my textbook on derivatives of products and quotients. Since I haven't used any derivatives of products or quotients in my solution to this exercise yet, I think I must be on the wrong track.
