Find the locus of the middle point of focal chord of parabola whose eq. is Y^2=4aX(standard eq.)

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\mathrm{y}}^2=4\mathrm{ax} \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{ends} \mathrm{of} \mathrm{focal} \mathrm{chord} \mathrm{are} \left({\mathrm{at}}^2,2\mathrm{at}\right) \mathrm{and} \left(\frac{\mathrm{a}}{{\mathrm{t}}^2},-\frac{2\mathrm{a}}{\mathrm{t}}\right) \\ \mathrm{Let} \left(\mathrm{h},\mathrm{k}\right) \mathrm{be} \mathrm{the} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{the} \mathrm{chord}. \\ \Rightarrow \mathrm{k}=\frac{2\mathrm{at}+\left(-{\displaystyle \frac{2\mathrm{a}}{\mathrm{t}}}\right)}{2} \\ \Rightarrow 2\mathrm{k}=2\mathrm{at}-\frac{2\mathrm{a}}{\mathrm{t}} \\ \Rightarrow 2\mathrm{k}=2\mathrm{a}\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right) \\ \Rightarrow \left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)=\frac{\mathrm{k}}{\mathrm{a}} ;\left(\mathrm{i}\right) \\ \mathrm{h}=\frac{{\mathrm{at}}^2+{\displaystyle \frac{\mathrm{a}}{{\mathrm{t}}^2}}}{2} \\ \Rightarrow 2\mathrm{h}={\mathrm{at}}^2+\frac{\mathrm{a}}{{\mathrm{t}}^2} \\ \Rightarrow \frac{2\mathrm{h}}{\mathrm{a}}={\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2} \\ \Rightarrow \frac{2\mathrm{h}}{\mathrm{a}}={\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)}^2+2 \left[{\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}={\left(\mathrm{a}-\mathrm{b}\right)}^2\right] \\ \Rightarrow \frac{2\mathrm{h}}{\mathrm{a}}={\left(\frac{\mathrm{k}}{\mathrm{a}}\right)}^2+2 \\ \Rightarrow \frac{2\mathrm{h}}{\mathrm{a}}=\frac{{\mathrm{k}}^2}{{\mathrm{a}}^2}+2 \\ \Rightarrow 2\mathrm{ah}={\mathrm{k}}^2+2{\mathrm{a}}^2 \\ \mathrm{To} \mathrm{get} \mathrm{equation} \mathrm{of} \mathrm{locus}, \mathrm{replace} \\ \mathrm{h}\to \mathrm{x} \mathrm{and} \mathrm{k}\to \mathrm{y} \\ \Rightarrow 2\mathrm{ax}={\mathrm{y}}^2+2{\mathrm{a}}^2 \\ \Rightarrow \boxed{{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{a}\left(\mathbf{x}\mathbf{-}\mathbf{a}\right)} \end{gathered}$$

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