Search any question & find its solution
Question:
Answered & Verified by Expert
The locus of the points of intersection of perpendicular normals to the parabola \(y^2=4 a x\) is
Options:
Solution:
1785 Upvotes
Verified Answer
The correct answer is:
\(y^2-a x+3 a^2=0\)
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 \(m_1 m_2=-1\) due to perpendicular normals.
\(\begin{aligned} & \text { so } m_3=\frac{k}{a} \\ & k=\frac{k}{a} h-2 a \frac{k}{a}-a\left(\frac{k}{a}\right)^3 \\ & \Rightarrow \mathrm{I}=\frac{h}{a}-2-\frac{k^2}{a^2} \\ & \end{aligned}\)
\(\Rightarrow \quad \frac{k^2}{a^2}=\frac{h}{a}-3 \Rightarrow k^2=a(h-3 a)\)
On taking locus \((h, k)\), we get
\(y^2-a x+3 a^2=0\)
Hence, option (4) is correct.
\(y=m x-2 a m-a m^3\) 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 \(m_1 m_2=-1\) due to perpendicular normals.
\(\begin{aligned} & \text { so } m_3=\frac{k}{a} \\ & k=\frac{k}{a} h-2 a \frac{k}{a}-a\left(\frac{k}{a}\right)^3 \\ & \Rightarrow \mathrm{I}=\frac{h}{a}-2-\frac{k^2}{a^2} \\ & \end{aligned}\)
\(\Rightarrow \quad \frac{k^2}{a^2}=\frac{h}{a}-3 \Rightarrow k^2=a(h-3 a)\)
On taking locus \((h, k)\), we get
\(y^2-a x+3 a^2=0\)
Hence, option (4) is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.