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Question

The minimum distance between the parabolas y24x8y+40=0 and x28x4y+40=0 is 


Solution

The equations of the parabolas are y24x8y+40=0(1)
x28x4y+40=0(2)
We observe that if we interchange x and y in equation (1), we obtain equation (2)
So, the two parabolas are symmetric about y=x.
Shortest distance exist between the tangents on both the parabolas which are parallel to y=x

For curve 1, the equation of tangent is
yy12(x+x1)4(y+y1)+40=0
2x+(y14)y2x14y1+40=0
Slope =2y14=1
y1=6
Putting value of y1 in eqn (1), we get x1=7
So, (x1,y1)=(7,6)
Since, the parabolas are symmetric about y=x, and if (x1,y1)=(7,6), then (x2,y2)=(6,7)
(x and y will interchange)

d=(x1x2)2+(y1y2)2=2 

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