Let A be the area between co-ordinate axis, y2=x−1,x2=y−1 and the line which makes the shortest distance between two parabolas and A be the area between x=0,x2=y−1,x=y and the shortest distance between y2=x−1andx2=y−1,
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Answer:
Notice that these are inverse functions of each other, you can swap x,y to get to the second parabola.
They are mirror images with respect to line x=y.
Required point should have this slope y′=1 for its tangent at point of tangency at ends of common normal.
Take the parabola with its symmetry axis coinciding with axis.
⇒y
2
=x−1
Differentiating w.r.t x we get,
⇒2yy′=1
⇒2y=1
⇒y=
2
1
⇒(
2
1
)
2
=x−1
⇒x=
4
5
Hence the x,y coordinates are
⇒(x,y)=(
4
5
,
2
1
)=(x
1
,y
1
) (say)
and the other point of tangency is again swapped to
⇒(x
2
,y
2
)=(
2
1
,
4
5
)
Now use distance formula between them to get the minimum distance
⇒d=
(
4
5
−
2
1
)
2
+(
2
1
−
4
5
)
2
⇒d=
(
4
3
)
2
+(
4
−3
)
2
⇒d=
4
3
2
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