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