Solution:
(d) Construct a function that describes the distance between ( x, y) and (4, 1).
By the “distance formula”, the distance, d, between ( x, y) and (4, 1) is
d=√(x−4)2+(y−1)2
For any point ( x, y) on the parabola, we have y = x2 +1,
Substituting for y in the formula for d, we have
d=√(x−4)2+([x2+1]−1)2
Simplifying, d=√(x−4)2+(x2)2=√(x−4)2+x4
Using the trace from function we can see that the minimum value of d occurs at (approximately) the point
x, d) = (1.12817, 3.4123)
That i.e., d is minimum when x is approximately 1.12817
The shortest distance between B and the parabola is approximately 3.14123
Using the equation for the parabola,y = x2 + 1 and the value of x that minimized d, ( x = 1.12817), we have y = 2.27277 (approximately).
So, the nearest point on the parabola y = x2 + 1, to the point (4, 1) is approximately (1.12817, 2.27277).
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