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, About the author Skylar

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 Reply

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