The shortest distance between the parabolas `2y^2=2x-1` and `2x^2=2y-1` is `2sqrt(2)` (b) `1/2sqrt(2)` (c) 4 (d) `sqrt((36)/5)`

Explanation

We observe that one of the equations of parabola is obtained by interchanging x and y in the other equation of parabola.
So, given parabolas are symmetrical about the line y=x.
For parabola, `y^(2)=(x-1//2)`, vertex is `(1//2,0)`
Now, we know that shortest distance occurs along common normal. Let common normal line meet parabolas at points A and B.
So, tangents to parabolas at oints A and B are parabola and they are also parabola to the line y=x.

Differentiating `2y^(2)=2x-1` w.r.t x, we get
`2y(dy)/(dx)=1`
`or " "(dy)/(dx)=(1)/(2y)=1` (since tangent is parallel to y=x)
`rArr" "y=(1)/(2)`
`So," "x=(3)/(4)`
`So," "A-=((3)/(4),(1)/(2))andB-=((1)/(2),(3)/(4))`
Hence, rquired shortest distance,
`AB=sqrt(((3)/(4)-(1)/(2))^(2)+((1)/(2)-(3)/(4))^(2))`
`=sqrt((1)/(16)+(1)/(16))=(1)/(2sqrt(2))`