A point P moves such that sum of the slopes of the normals drawn from it to the hyperbola xy=16 is equal to the sum of ordinates of feet of normals. The locus of P is a curve C

Options

A. `x^(2)=4y`
B. `x^(2)=16y`
C. `x^(2)=12y`
D. `y^(2)=8x`

Answer

B

Explanation

Any point on the hyperbola xy = 16 is `(4t, 4//t)`.
Normal at this point is `y-4//t=t^(2)(x-4t)`.
If the normal passes through P(h,k), then0
`k-(4)/(t)=r^(2)(h-4t)`
`"or "4t^(4)-t^(3)h+tk-4=0`
The equation has roots `t_(1),t_(2),t_(3),t_(4)` which are parameters of the four feet of normals on the hyprbola. Therefore,
`sumt_(1)=(h)/(4)`
`sumt_(1)t_(2)=0`
`sumt_(1)t_(2)t_(3)=-(k)/(4)`
`"and "t_(1)t_(2)t_(3)t_(4)=-1`
`therefore" "(1)/(t_(1))+(1)/(t_(2))+(1)/(t_(3))+(1)/(t_(4))=(k)/(4)`
`"or "y_(1)+y_(2)+y_(3)+y_(4)=k`
According to the question,
`t_(1)^(2)+t_(2)^(2)+t_(3)^(2)+t_(4)^(2)=(h^(2))/(16)=k`
Hence, the locus of (h,k) is
`x^(2)=16y`