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The locus of the foot of the perpendicul...

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`

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`

Text Solution

Verified by Experts

The correct Answer is:
D
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`
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