The shortest distance between the line x=y and the curve `y^(2)=x-2` is
(a) 2 (b) `(7)/(8)` (c) `(7)/(4sqrt(2))` (d) `(11)/(4sqrt(2))`

Options

A. 2
B. `(7)/(8)`
C. `(7)/(4sqrt(2))`
D. `(11)/(4sqrt(2))`

Answer

C

Explanation

Given equation of curve is
`y^(2)x-2" "...(i)`
and the equation of line is
`y=x" "...(ii)`

Consider a point `P(t^(2)+2,t)` on parabola (i).
For the shortest distance between curve (i) and line (ii), the line PM should be perpendicular to line (ii) and parabola, (i). i.ee, tangent at P should be paralle to y=x
`:. (dx)/(dx)|_(at "poitns P")` = Slope of tangent at point P to curve (i)
`=1 " " [ :. "tangent is parallel"]`
`rArr (1)/(2y)|_(p)=1" "` [ differentiating the curve (i), we get `2y(dy)/(dx)=1`]
`rArr (1)/(2t) =1 rArr t=(1)/(2)" "[ :.P(x,y) =P(t^(2)+2,t)]`
Now, the point P is `((9)/(4),(1)/(2))`.
Now, minimum distane = `PM=(|(9)/(4)-(1)/(2)|)/(sqrt(2))`
[ `:.` distane of a point `P(x_(1),y_(1))` from a line `ax+by +c =0 is (|ax_(1)+by_(1)+c|)/(sqrt(a^(2)+b^(2)))]`
`=(7)/(4sqrt(2))` units