AP EAMCET · Maths · Parabola
The locus of the points of intersection of perpendicular normals to the parabola \(y^2=4 a x\) is
- A \(y^2-2 a x+a^2=0\)
- B \(y^2+a x+2 a^2=0\)
- C \(y^2-a x+2 a^2=0\)
- D \(y^2-a x+3 a^2=0\)
Answer & Solution
Correct Answer
(D) \(y^2-a x+3 a^2=0\)
Step-by-step Solution
Detailed explanation
Let the equation of normal having slope ' \(m\) ' to the parabola \(y^2=4 a x\) is \(y=m x-2 a m-a m^3\) and passes through \((h, k)\), then \(k=m h-2 a m-a m^3\) is cubic equation in ' \(m\) '. Let having roots \(m_1, m_2\) and \(m_3\). so, \(m_1 m_2 m_3=-\frac{k}{a}\) and…
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