JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
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 \(L_{1}\) and \(L_{2}\) meet on the straight line
- A \(x+3=0\)
- B \(x+2 y=0\)
- C \(2 x+1=0\)
- D \(x+2=0\)
Answer & Solution
Correct Answer
(A) \(x+3=0\)
Step-by-step Solution
Detailed explanation
\(y^{2}=4(x+1)\) equation of tangent \(y=m(x+1)+\frac{1}{m}\) \(y=m x+m+\frac{1}{m}\) \(y^{2}=8(x+2)\) equation of tangent \(y = m ^{\prime}( x +2)+\frac{2}{ m ^{\prime}}\) \(y = m ^{\prime} x +2\left( m ^{\prime}+\frac{1}{ m ^{\prime}}\right)\) since lines intersect at right…
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