Let $\mathrm{L}_{1}$ be a tangent to the parabola $y^{2}=4(x+1)$ and $\mathrm{L}_{2}$ be a tangent to the parabola $y^{2}=8(x+2)$ such that $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ intersect at right angles. Then $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ meet on the straight line:
Correct Option: 1
$L_{1}: y=m_{1}(x+1)+\frac{1}{m_{1}} \quad$ [Tangent to $\left.y^{2}=4(x+1)\right]$
$L_{2}: y=m_{2}(x+2)+\frac{2}{m_{2}} \quad$ [Tangent to $y^{2}=8(x+2)$ ]
$m_{1}^{2}(x+1)-y m_{1}+1=0$
$m_{2}^{2}(x+2)-y m_{2}+2=0$
$\because m_{2}=-\frac{1}{m_{1}} \quad\left(\because L_{1} \perp L_{2}\right)$
[From(ii)]
$\Rightarrow 2 m_{1}^{2}+y m_{1}+(x+2)=0$ (iii)
From (i) and (iii),
$\frac{x+1}{2}=\frac{-y}{y}=\frac{1}{x+2} \Rightarrow x+3=0$
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