The locus of the points of intersection of perpendicular normals to the parabola \(y^2=4 a x\) is

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.