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