Minimum distance between the parabola `y^2-4x-8y+40=0" and "x^2-8x-4y+40=0` is

Options

A. 0
B. `sqrt3`
C. `2sqrt2`
D. `sqrt2`

Answer

D

Explanation

The equations of the parabolas are
`y^(2)-4x-8y+40=0" ....(i)"`
`x^(2)-8x-4y+40=0" ....(ii)"`
We observe that if we interchange x and y in equation (i), we obtain equation (ii), So, the two parabolas are symmetric about y = x.
The two parabolas intersect on y = x, then the minimum distance between them is zero.
On solving y = x and `y^(2)-4x-8y+4=0`, we get
`x^(2)-12x+40=0`, which has imaginary roots.
So, the parabolas do not intersect. Consequently, distance between them is not zero.
Minimum distance between the two parabolas in the distance between tangents to the two parabolas which are parallel to y = x.
Differentiatiing (ii), w.r. to x, we get
`2x-8-4("dy")/("dx")=0rArr("dy")/("dx")=(x-4)/2`
`(x-4)/2=1rArrx=6`
Putting x = 6 in (ii), we get y = 7.
Thus, the coordinates of a point on parabola (ii), are P(6, 7). The corresponding point on parabola (i) is `Q(7, 6).`
Hence, requirex distance = `PQ=sqrt(1+1)=sqrt2`.