Find the value of a for which the area of triangle formed by the tangent and the coordinate axes will be minimum.
Equation of tanget at P : y = a^2 - 2ax + 4
Tangent : P (a , 4-a^2)
Thanks
Equation of line (assuming it is correct):
y = a
2-2ax+4
y-intercept occurs at y = 4
x-intercept occurs at x = (a
2+4)/2a
Hence area of triangle is:
A = 1/2 (4) (a
2+4)/2a
= (a
2+4)/a
= a + 4/a
Want to minimise A (> 0) with respect to a
dA/da = 1 - 4/a
2
dA/da = 0 => a = 2 or - 2
d
2A/da
2 = 8/a
3
The second derivative is positive (for a minimum) when a > 0, hence choose a = 2 which gives a local minimum. Since A > 0, then domain is: a > 0 thus there are no other turning points in this domain, so the local minimum is also the absolute minimum.