The locus of the foot of the perpendicular from the centre of the hyperbola `xy = c^2` on a variable tangent is (A) `(x^2-y^2)=4c^2xy` (B) `(x^2+y^2)^2=2c^2xy` (C) `(x^2+y^2)=4c^2xy` (D) `(x^2+y^2)^2=4c^2xy`

Options

A. `(x^(2)-y^(2))^(2) = 4c^(2)xy`
B. `(x^(2)+y^(2))^(2) =2c^(2)xy`
C. `(x^(2)-y^(2))^(2)=2c^(2)xy`
D. `(x^(2)+y^(2))^(2)=4c^(2)xy`

Answer

D

Explanation

Equation of tangent at `P(ct,c//t)` is `(x)/(t) + ty = 2c` or `x + t^(2) y = 2ct` (1)
Slope of tangent `=- (1)/(t^(2))`
`:.` Equation of perpendicular to this tangent from (0,0) is `y = t^(2)x`
Squaring (i), `(x+ t^(2)y)^(2) = 4c^(2)t^(2)`
Putting the value of `t^(2)` from (2) in above we get
`(x+(y^(2))/(x))^(2) = 4c^(2) (y)/(x)`
`rArr (x^(2)+y^(2)) = 4c^(2)xy`