Consider the parabola y^2=4x. Let S be the fo | vedclass

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(B, C) The equation of a tangent to the parabola $y^2=4x$ (where $a=1$) is $y=mx+\frac{1}{m}$.Since it passes through $P=(-2,1)$,we have $1=-2m+\frac{

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$B, C, D$

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(B, C) The equation of a tangent to the parabola $y^2=4x$ (where $a=1$) is $y=mx+\frac{1}{m}$.Since it passes through $P=(-2,1)$,we have $1=-2m+\frac{1}{m}$,which simplifies to $2m^2+m-1=0$.Solving for $m$,we get $(2m-1)(m+1)=0$,so $m=\frac{1}{2}$ or $m=-1$.The points of contact are given by $(\frac{a}{m^2}, \frac{2a}{m})$.For $m=\frac{1}{2}$,the point is $P_1=(4,4)$. For $m=-1$,the point is $P_2=(1,-2)$.The focus $S$ is $(1,0)$.The line $SP_1$ passes through $(1,0)$ and $(4,4)$,so its equation is $y-0=\frac{4-0}{4-1}(x-1)$,which is $4x-3y-4=0$.The line $SP_2$ passes through $(1,0)$ and $(1,-2)$,so its equation is $x=1$.The length $PQ_1$ is the perpendicular distance from $P(-2,1)$ to $4x-3y-4=0$,which is $PQ_1 = \frac{|4(-2)-3(1)-4|}{\sqrt{4^2+(-3)^2}} = \frac{|-15|}{5} = 3$.Similarly,$PQ_2$ is the perpendicular distance from $P(-2,1)$ to $x=1$,which is $PQ_2 = |-2-1| = 3$.In $	riangle SPQ_1$,$SQ_1 = \sqrt{SP^2 - PQ_1^2}$. Here $SP = \sqrt{(-2-1)^2+(1-0)^2} = \sqrt{9+1} = \sqrt{10}$.So $SQ_1 = \sqrt{10-9} = 1$. Similarly,$SQ_2 = 1$.Thus,options $(B)$ and $(C)$ are correct.

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