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

Answers (1)

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

Correct Answer: Option C

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