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What is the shortest distance between the circle x^2 + (y+6)^2 = 1 and the parabola y^2 = 8x?

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What is the shortest distance between the circle x^2 + (y+6)^2 = 1 and the parabola y^2 = 8x?
posted Sep 26 by anonymous

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1 Answer

–1 vote

Minimum distance between any two curves lie along common normal
So if we find the coordinate where common normal touches the parabola, then that coordinate will be closest to circle
y^2 = 8x
x^2 + (y+6)^2 = 1
Equation normal to parabola:
y = mx - 2(2)m-2m^3
where m is slope of normal
y = mx - 4m-2m^3
So this common normal will pass through the center of circle also
Center (0, -6)
So, -6 = m(0) - 4m-2m^3
m^3 + 2m - 3 = 0
m = 1 is one solution
y = x - 6
Put this in parabola
y^2 = 8(y+6)
y^2 - 8y - 48 = 0
y = -4, 12
x = 2, 18
(2, -4) is the closest point

answer Oct 6 by Hanifa Mammadov



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