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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 intervalA. `(1,4)`B. `(-oo,0)uu(4,oo)`C. `(-4,4)`D. `[3,4]` |
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Answer» Correct Answer - 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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