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1) dy / dx (1 / x) = -1/x^2
2) the slope of the tangent at "a" = -1/a^2
3) Equation of the tangent =
y - 1/a = ( -1 /a^2) ( x - a)
y = -x / a^2 + 2/a
y = [ 2a - x ] / a^2
B. I think you want to know where the tangent line intersects the y and x axis.
Wnen x = 0, y = 2/a this is the y intercept = (0, 2/a)
When y = 0, x = 2a this is the x intercept = (2a, 0 )
C. I think you want to find the area of a triangle bounded by the tangent line and both axis....if so, we have......
A = (1/2)(b)(h) = (1/2)(2a)(2/a) = 4 / 2 = 2 sq units........this is a constant area that does NOT depend upon "a"
D. We have
sqrt ( [2/a]^2 + [2a]^2 ) = 5 square both sides
4/a^2 + 4a^2 = 25 multiply through by a^2
4 + 4a^4 = 25a^2 rearrange
4a^4 - 25a^2 + 4 = 0
In the first quadrant, we have two solutions ... a ≈ .40536 and a ≈ 2.4669
So we have ( .40536, 1/.40536) and (2.4669, and 1/ 2.4669 )
Here's a graph of both tangent line solutions......note that the total length of both tangent lines between their intersections with the x and y axis ≈ 5
https://www.desmos.com/calculator/km8gajbnnl
