Let `vecr` be a position vector of a variable point in Cartesian OXY plane such that `vecr. ( 10 hatj - 8 hati-vecr) =40` and `P_(1)=max {|vecr+2hati - 3hatj|^(2)} , P_(2) = min {|vecr+ 2hati - 3 hatj|^(2)}` . A tangenty line is drawn to the curve `y= 8// x^(2)` at point .A with abscissa 2. the drawn line cuts the x-axis at a point B.
`p_(2)` is equal to

Options

A. 2
B. 10
C. 18
D. 5

Answer

c

Explanation

Let `vecx= x hati + yhatj`
`x^(2) + y^(2) + 8x - 10y + 40 =0` , which is a circle
centre C(-4,5) , radius r = 1
`p_(1)= max {(x+2)^(2)+ (y-3)^(2)}`
`P_(2) = min {(x+2)^(2)+ (y-3)^(2)}`
Let P be (-2,3). Then
`CP = sqrt2,r=1`
` P_(2)= (2sqrt2-1)^(2)`
`P_(1) = (2sqrt2+1)^(2)`
`P_(1) + p_(2) =18`
Slope = AB = `((dy)/(dx))_(2,2)=-2`
Equation of AB, 2x+y=6
`vec(OA)=2hati=2hatj,vec(OB)= 3hati`
`vec(AB)=hati-2hatj`
`vec(AB).vec(OB)= (hati-2hatj) (3hati)=3`