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The coordinate of the point on y^2=8x which is closest from x^2 + (y+6)^2 =1 is ?

  1. The coordinate of the point on y^2=8x which is closest from x^2 + (y+6)^2 =1 is ?

Grade:12th pass

3 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

Minimum distance b/w any two curves lie along common normal
So if we find the coordinate where common normal touches the parabola, then thats 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 centre of circle also
Centre (0, -6)
So,
-6 = m(0) - 4m-2m^3
m^3 + 2m - 3 = 0
m = 1 is one solution which is also clear from the figure.
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 which is clear from the diagram.
Arpit Dhankar
33 Points
9 years ago
thank you sir
Arpit Dhankar
33 Points
9 years ago
thank you sir

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