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    				| 1. | The equation to the line touching both the parabolas `y^2 =4x` and `x^2=-32y` isA. `x+2y+4=0`B. `2x+y-4=0`C. `x-2y-4=0`D. `x-2y+4=0` | 
| Answer» Correct Answer - A The equation of any tangent to the parabola `y^(2)=4x` is `y=mx+(1)/(m)....(i)` This touches the parabola `x^(2)=-32y`, therefore the equation `x^(2)=-32(mx+(1)/(m))` has equal roots `therefore (32m)^(2)=4((32)/(m))` `=Rightarrow 8m^(3)=1 Rightarrow m=(1)/(2)` On putting the value of m in Eq. (i), we get x-2y+4=0 | |