The midpoints of focal chords of y^2 = 16x is?  [Ans: y^2 = 8(x-4)]

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$$\begin{gathered} \mathrm{I} \mathrm{assume} \mathrm{you} \mathrm{need} \mathrm{locus} \mathrm{of} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{focal} \mathrm{chord}. \\ \mathrm{We} \mathrm{have} \\ {\mathrm{y}}^2=16\mathrm{x} \\ \mathrm{Compare} \mathrm{with} {\mathrm{y}}^2=4\mathrm{ax} \\ \Rightarrow 4\mathrm{a}=16 \\ \Rightarrow \mathrm{a}=4 \\ \mathrm{Ends} \mathrm{of} \mathrm{focal} \mathrm{chord} \mathrm{are} \mathrm{A}\left({\mathrm{at}}^2,2\mathrm{at}\right) \mathrm{and} \mathrm{B}\left(\frac{\mathrm{a}}{{\mathrm{t}}^2},-\frac{2\mathrm{a}}{\mathrm{t}}\right) \\ \Rightarrow \mathrm{A}\left(4{\mathrm{t}}^2,8\mathrm{t}\right) \mathrm{and} \mathrm{B}\left(\frac{4}{{\mathrm{t}}^2},-\frac{8}{\mathrm{t}}\right) \\ \mathrm{Let} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{focal} \mathrm{chord} \mathrm{be} \left(\mathrm{h},\mathrm{k}\right) \\ 2\mathrm{k}=8\mathrm{t}-\frac{8}{\mathrm{t}} \\ \Rightarrow 2\mathrm{k}=8\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right) \\ \Rightarrow \frac{\mathrm{k}}{4}=\mathrm{t}-\frac{1}{\mathrm{t}}...\left(\mathrm{i}\right) \\ \Rightarrow 2\mathrm{h}=4{\mathrm{t}}^2+\frac{4}{{\mathrm{t}}^2} \\ \Rightarrow 2\mathrm{h}=4\left({\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}\right) \\ \Rightarrow \frac{\mathrm{h}}{2}=\left({\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}\right) \\ \Rightarrow \frac{\mathrm{h}}{2}={\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)}^2+2 \\ \mathrm{Use} \left(\mathrm{i}\right) \\ \frac{\mathrm{h}}{2}={\left(\frac{\mathrm{k}}{4}\right)}^2+2 \\ \Rightarrow \frac{\mathrm{h}}{2}=\frac{{\mathrm{k}}^2}{16}+2 \\ \Rightarrow \mathrm{h}=\frac{{\mathrm{k}}^2}{8}+4 \\ \Rightarrow \frac{{\mathrm{k}}^2}{8}=\mathrm{h}-4 \\ \Rightarrow {\mathrm{k}}^2=8\left(\mathrm{h}-4\right) \\ \mathrm{Hence} \mathrm{locus} \mathrm{is} \\ {\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{8}\left(\mathbf{x}\mathbf{-}\mathbf{4}\right) \end{gathered}$$

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