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Consider a hyperbola xy = 4 and a line y...

Consider a hyperbola `xy = 4` and a line `y = 2x = 4`. O is the centre of hyperbola. Tangent at any point P of hyperbola intersect the coordinate axes at A and B.
Locus of circumcentre of triangle OAB is

A

an ellipse with eccentricity `(1)/(sqrt(2))`

B

a parabola

C

a hyperbola with eccnetricity `sqrt(2)`

D

a circle

Text Solution

Verified by Experts

The correct Answer is:
C
Let `(2t,2//t)` be a point on the hyperbola. Equation of the tangent at this point `x + yt^(2) = 4t`.
`:. A = (4t,0),B = (0,4//t)`
Locus of circumcentre of triangle is `xy = 16`
Its eccentricity is `sqrt(2)`
Shortest distance exist along the common normal.
`:. t^(2) = 1//2` or `t = 1//sqrt(2)`
`:.` Foot of the perpendicular is `C (sqrt(2),2sqrt(2))`
`:.` Shortest distance is distance of C from the given line which is `(4(sqrt(2)-1))/(sqrt(5))`
Given line intersect the x-axis at `R(2,0)` Any point on this line at distance 'r' from R is `(2+r cos theta, r sin theta)` If this point lies on hyperbola, then we have `(2+r cos theta) (r sin theta) =4`
Product of roots of above quadratie in 'r' is `r_(1)r_(2) = 8//|sin 2 theta|`, which has minimum value 8
`:.` Minimum value of `RS xx RT` is 8
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