Posted by Lynnn on Sunday, April 8, 2012 at 6:08pm.
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
or
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
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