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The slope of the line touching both the parabolas `y^2=4x and x^2=−32y` is (a) `1/2` (b) `3/2` (c) `1/8` (d) `2/3`A. 43473B. 43499C. 43467D. 43526 |
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Answer» Correct Answer - C The equation of any tangent to `y^(2)=4x` is `y=mx+1/m"or, "x=1/my-1/m^(2)`. If it touches `x^(2)=-32y,` then As the equation of any tangent to `x^(2)=-32y`, is of the form `x=my-8/m.` `:." "-1/m^(2)=-8/("1/m")rArrm=1/2` ALITER The equation of any tangent to `y^(2)=4x" is "y =mx+1/m`. If it touches `x^(2)=-32y," then "x^(2)=-32(mx+1/2)` must have equal roots. `:." "(32m)^(2)=4((32)/(m))rArr=m=1/2` |
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