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The value of parameter t so that the lin...

The value of parameter t so that the line `(4-t)x+ty+(a^(3)-1)=0` is normal to the curve xy = 1 may lie in the interval

A

`(1,4)`

B

`(-oo,0)uu(4,oo)`

C

`(-4,4)`

D

`[3,4]`

Text Solution

Verified by Experts

The correct Answer is:
B
Slope of line `(4-t)x+ty+(a^(3)-1)=0" is "(t-4)/(t)`
For `xy=1, (dy)/(dx)=(-y)/(x)=(-1)/(x^(2))`
`therefore" Slope of normal "=-x^(2)=(t-4)/(t)`
As `x^(2)gt0,(t-4)/(t)gt0`
`therefore" "t in (-oo,0)uu(4,oo)`
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