Let L be a normal to the parabola y^2=4x. If | vedclass

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(C) The equation of a normal to the parabola $y^2=4ax$ (where $a=1$) is given by $y=mx-2am-am^3$.Substituting $a=1$,we get $y=mx-2m-m^3$.Since the nor

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$(A, B, D)$

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(C) The equation of a normal to the parabola $y^2=4ax$ (where $a=1$) is given by $y=mx-2am-am^3$.Substituting $a=1$,we get $y=mx-2m-m^3$.Since the normal passes through $(9,6)$,we have:$6 = 9m - 2m - m^3$$6 = 7m - m^3$$m^3 - 7m + 6 = 0$By testing values,we find $m=1$ is a root. Dividing by $(m-1)$,we get $(m-1)(m^2+m-6)=0$,which factors to $(m-1)(m+3)(m-2)=0$.Thus,the possible slopes are $m=1, m=2, m=-3$.For $m=1$: $y = 1x - 2(1) - (1)^3$ $\Rightarrow y = x - 3$ $\Rightarrow y-x+3=0$.For $m=2$: $y = 2x - 2(2) - (2)^3$ $\Rightarrow y = 2x - 4 - 8$ $\Rightarrow y-2x+12=0$.For $m=-3$: $y = -3x - 2(-3) - (-3)^3$ $\Rightarrow y = -3x + 6 + 27$ $\Rightarrow y+3x-33=0$.The equations are $y-x+3=0$,$y-2x+12=0$,and $y+3x-33=0$. These correspond to options $(A)$,$(B)$,and $(D)$.

Let $L$ be a normal to the parabola $y^2=4x$. If $L$ passes through the point $(9,6)$,then $L$ is given by
$(A)$ $y-x+3=0$ $(B)$ $y+3x-33=0$ $(C)$ $y+x-15=0$ $(D)$ $y-2x+12=0$

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Let $L$ be a normal to the parabola $y^2=4x$. If $L$ passes through the point $(9,6)$,then $L$ is given by$(A)$ $y-x+3=0$ $(B)$ $y+3x-33=0$ $(C)$ $y+x-15=0$ $(D)$ $y-2x+12=0$