Let $L_{1}$ be a tangent to the parabola $y^{2}=4(x+1)$ and $L_{2}$ be a tangent to the parabola $y^{2}=8(x+2)$ such that $L_{1}$ and $L_{2}$ intersect at right angles. Then $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ meet on the straight line :
Correct Option: 1
$y^{2}=4(x+1)$
equation of tangent $\mathrm{y}=\mathrm{m}(\mathrm{x}+1)+\frac{1}{\mathrm{~m}}$
$y=m x+m+\frac{1}{m}$
$y^{2}=8(x+2)$
equation of tangent $\mathrm{y}=\mathrm{m}^{\prime}(\mathrm{x}+2)+\frac{2}{\mathrm{~m}^{\prime}}$
$\mathrm{y}=\mathrm{m}^{\prime} \mathrm{x}+2\left(\mathrm{~m}^{\prime}+\frac{1}{\mathrm{~m}^{\prime}}\right)$
since lines intersect at right angles
$\therefore \mathrm{mm}^{\prime}=-1$
Now $y=m x+m+\frac{1}{m}$ $\ldots(1)$
$y=m^{\prime} x+2\left(m^{\prime}+\frac{1}{m^{\prime}}\right)$
$y=-\frac{1}{m} x+2\left(-\frac{1}{m}-m\right)$
$\mathrm{y}=-\frac{1}{\mathrm{~m}} \mathrm{x}-2\left(\mathrm{~m}+\frac{1}{\mathrm{~m}}\right)$ $\ldots(2)$
From equation (1) and (2)
$m x+m+\frac{1}{m}=-\frac{1}{m} x-2\left(m+\frac{1}{m}\right)$
$\left(\mathrm{m}+\frac{1}{\mathrm{~m}}\right) \mathrm{x}+3\left(\mathrm{~m}+\frac{1}{\mathrm{~m}}\right)=0$
$\therefore x+3=0$
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