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If a point P is taken on xy=2 and then a...

If a point `P` is taken on `xy=2` and then a normal is drawn from `P` on the ellipse `(x^(2))/6+(y^(2))/3=1` which is perpendicular to `x+y=8`, then `P` is

A

`(1,2)`

B

`(-1,-2)`

C

`(2,1)`

D

`(-2,-1)`

Text Solution

Verified by Experts

The correct Answer is:
B, C

`:' sqrt(6)xsectheta-sqrt(3)ycosectheta=3` is equation of normal,
then, slope `=(-sqrt(6)sectheta)/(-sqrt(3)cosectheta)=sqrt(2) tan theta`
Given, `sqrt(2)tantheta=+1`
`impliestantheta=1/(sqrt(2))`
`:.` normal is `x-y=1`
Let `P(h, 2/h)`
`implies h-2/h=1`
`impliesh^(2)h-2=0`
`impliesh=2` or `-1`
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