Given parabola y² = 8x
and the point (a,0).
Let P(x, y) be any point on the parabola.
d=(x−a)2+y2
D=d2=(x−a)2+y2
D(x)=(x−a)2+8x
Differentiate with respect to x
D′(x)=2(x−a)+8
D′(x)=0
2(x−a)+8=0
x−a=−4
x=−4+a
D′(x)=2(x−a)+8
Differentiate with respect to x
D′′(x)=2x
D′′(−4+a)=2(−4+a)=−8+a
If a > 8 then D"(-4+a) >0
The minimum distance is
(−4+a−a)2+8(−4+a)
=16−12+8a
= 4+8a
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