Let `vec(r)` is a positive vector of a variable pont in cartesian OXY plane such that `vecr.(10hatj-8hati-vecr)=40` and `p_1=max{|vecr+2hati-3hatj|^2},p_2=min{|vecr+2hati-3hatj|^2}`. A tangent line is drawn to the curve `y=8/x^2` at the point A with abscissa 2. The drawn line cuts x-axis at a point B

Options

A. 9
B. `2sqrt2-1`
C. `6sqrt6+3`
D. `9-4sqrt2`

Answer

d

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`