Solve this following


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 :

  1. $x+3=0$

  2. $x+2 y=0$

  3. $2 x+1=0$

  4. $x+2=0$

Correct Option: 1



equation of tangent $\mathrm{y}=\mathrm{m}(\mathrm{x}+1)+\frac{1}{\mathrm{~m}}$

$y=m x+m+\frac{1}{m}$


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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