posted by Lynnn on .
Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2+7 and connect the tangent point to the x-axis. If the tangent point is close to the y-axis, the line segment is long. If the tangent point is far from the y-axis, the line segment is also very long. Which tangent point has the shortest line segment?
(Suppose C is a positive number. What point on the curve has first coordinate equal to C?)
answer to C question at the end: If 1st coordinate (x-value) is C, the 2nd coordinate (y-value) is C^2+7.
Now, for the other question:
At (x,y), the slope of the tangent line is
dy/dx = 2x
So, now we have a point (a,a^2+7) and a slope (2a).
The equation of the tangent line where x=a is thus
y-(a^2+7) = 2a (x-a)
y = 2ax - 2a^2 + a^2+7
y = 2ax - a^2 + 7
This line intersects the x-axis where
0 = 2ax - a^2 + 7
x = (a^2-7)/2a
So, the length of the line segment from (a,a^2+7) to (0,(a^2-7)/2a) is
d^2 = (a - (a^2-7)/2a)^2 + (a^2+7)^2
d = 1/2a (a^2 + 7) sqrt(4a^2 + 1)
dd/da = (8a^4 + a^2 - 7)/[2a^2 sqrt(4a^2+1)]
dd/da = 0 when x = ±√(7/8) = ±0.935
d = 8.929