If two tangents to the parabola y^2=8x meet t | vedclass

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(C) Given parabola is $y^2=8x$. Comparing with $y^2=4ax$,we get $4a=8 \Rightarrow a=2$. Let the parametric coordinates of $P$ and $Q$ be $(at_1^2, 2at

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$y^2=8(x+2)$

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(C) Given parabola is $y^2=8x$. Comparing with $y^2=4ax$,we get $4a=8 \Rightarrow a=2$. Let the parametric coordinates of $P$ and $Q$ be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ respectively. So,the equation of the tangent at $P$ is $yt_1=x+at_1^2 \Rightarrow yt_1=x+2t_1^2$ $(i)$. Similarly,the equation of the tangent at $Q$ is $yt_2=x+2t_2^2$ $(ii)$. The tangent at the vertex of the parabola $y^2=8x$ is the $y$-axis,i.e.,$x=0$. To find $M$,we put $x=0$ in equation $(i)$,we get $yt_1=2t_1^2 \Rightarrow y=2t_1$. Thus,$M=(0, 2t_1)$. To find $N$,we put $x=0$ in equation $(ii)$,we get $yt_2=2t_2^2 \Rightarrow y=2t_2$. Thus,$N=(0, 2t_2)$. Given $MN=4$,so $|2t_1-2t_2|=4$ $\Rightarrow |t_1-t_2|=2$ $\Rightarrow (t_1-t_2)^2=4$ $(iii)$. The point of intersection $R(h, k)$ of the tangents at $P$ and $Q$ is given by $(at_1t_2, a(t_1+t_2))$. So,$(h, k) = (2t_1t_2, 2(t_1+t_2))$. This implies $t_1t_2 = h/2$ and $t_1+t_2 = k/2$. We know $(t_1+t_2)^2 = (t_1-t_2)^2 + 4t_1t_2$. Substituting the values,$(k/2)^2 = 4 + 4(h/2)$ $\Rightarrow k^2/4 = 4+2h$ $\Rightarrow k^2 = 16+8h = 8(h+2)$. Replacing $(h, k)$ with $(x, y)$,the locus is $y^2=8(x+2)$.

List-$I$	List-$II$
$(I)$ Vertex	$(A)$ $(-\frac{3}{2}, -3)$
$(II)$ Focus	$(B)$ $(\frac{3}{2}, -3)$
$(III)$ Equation of the directrix	$(C)$ $2x+5=0$
$(IV)$ Equation of the axis	$(D)$ $2x+y+3=0$
	$(E)$ $y+3=0$
	$(F)$ $(-2, -3)$

If two tangents to the parabola $y^2=8x$ meet the tangent at its vertex in $M$ and $N$ such that $MN=4$,then the locus of the point of intersection of those two tangents is

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