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The range of values of `alpha`for which the line `2y=gx+alpha`is a normal to the circle `x^2=y^2+2gx+2gy-2=0`for all values of `g`is`[1,oo)`(b) `[-1,oo)``(0,1)`(d) `(-oo,1]`A. `[1,oo)`B. `[-1,oo)`C. `(0,1)`D. `(- oo,1]` |
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Answer» Correct Answer - 2 The line `2y=g x +alpha` should pass through `(-g, -g)` . So, `-2g = -g^(2)+alpha` or `alpha= g^(2)-2g = (g-1)^(2) - ge -1` |
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