following the hints suggested:
let the point be (c,c^2 + 1)
dy/dx = 2x
so at (c,c^2+1) the slope of the tangent is 2c
let the tangent equation be y = mx + b
y = 2cx + b
for our point,
c^2 + 1 = 2c(c) + b
b = 1-c^2
so the tangent equation is
y = 2cx + 1-c^2
at the x-intercept,
0 = 2cx + 1-c^2
x = (c^2 - 1)/(2c)
then using the distance formula
D^2 = (c^2+1)^2 + (c - (c^2 - 1)/(2c))^2
Ok, I will now expand this. At first I thought to differenctiate using quotient rule for the last term, but it looked rather messy
D^2 = c^4 + 2c^2 + 1 + c^2 - c^2 + 1 + (c^4-2c^2+1)/(4c^2)
= c^4 + 2c^2 + 2 + (1/4)c^2 - 1/2 + (1/4)c^-2
2D(dD/dc) = 4c^3 + 4c + c/2 - 1/(2c^3) = 0 for a max/min of D
8c^6 + 9c^4 - 1 = 0
getting really messy....
let a = c^2
solve 8a^3 + 9a^2 - 1 = 0
a=-1 works !!!!!!
(a+1)(8a^2 + a - 1) = 0
if a=-1, c^=-1 ---> no solution
8a^2 + a - 1 = 0
a = (-1 ± √33)/16 = .2965 or a negative
c^2 = .2965
c = .5145
Please, please check my arithmetic and algebra, the method is right!
Thanks so much, but then do you plug in c back into the first point and tangent line and x-intercept to get the x and y coordinates and tangent line equation and x-intercept, respectively?
Can you explain to me how you found out the distance formula. Because how did you find out D^2 = c^4 + 2c^2 + 1 + c^2 - c^2 + 1 + (c^4-2c^2+1)/(4c^2)? I understand the first part with c^4 + 2c^2 + 1, but I don't understand how you got the second part.
And isn't (c^4-2c^2+1)/(4c^2) supposed to be (c^4+2c^2+1)/(4c^2)?
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