question_answer

The equation of a straight line drawn through the focus of the parabola \[{{y}^{2}}=-4x\] at an angle of 120° to the x-axis is [Orissa JEE 2005]

A)            \[y+\sqrt{3}(x-1)=0\]                

B)            \[y-\sqrt{3}(x-1)=0\]

C)            \[y+\sqrt{3}(x+1)=0\]               

D)            \[y-\sqrt{3}(x+1)=0\]

Correct Answer: C

Solution :

           \[m=\tan (120{}^\circ )=-\sqrt{3}\] = Slope of the line which passes                        through (?1, 0). Required equation, \[y-0=-\sqrt{3}(x+1)\] \[y+\sqrt{3}(x+1)=0\].