# Let  $L_{1}$ be the tangent to the parabola $y^2=4(x+1)$ and $L_{2}$   be the tangent to the parabola $y^2=8(x+2)$ such that $L_{1}$ and $L_{2}$ intersect at $90^{o}$. Then at $L_{1}$ and $L_{2}$ meet on the st. line.   Option: 1 2x+1=0    Option: 2 x+2=0    Option: 3 x+3=0      Option: 4 x+2y=0

Equation of 1st parabola :  $y^2=4(x+1)$

Equation of tangent : $y=m_1(x+1)+\frac{1}{m_1}$

Equation of 2nd parabola : $y^2=8(x+2)$

Equation of tangent : $y=m_2(x+2)+\frac{2}{m_2}$

Now both tangents intersect at 90 degree

$So,\;m_1m_2=-1$

$\\m_1(x+1)+\frac{1}{m_1}=m_2(x+2)+\frac{2}{m_2}\\x(m_1-m_2)+m_1-2m_2=\frac{2}{m_2}-\frac{1}{m_1}\\x(m_1-m_2)+m_1-2m_2=\frac{2m_1-m_2}{m_1m_2}\\x(m_1-m_2)+m_1-2m_2=\frac{2m_1-m_2}{-1}\\x(m_1-m_2)=3m_2-3m_1\\x+3=0$

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