TP and TQ are tangents to y2=4ax and the normals at p and q meet at a point r on the curve. find the locus of the center of the triangle tpq.

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$$\begin{gathered} \mathrm{To} \mathrm{solve} \mathrm{this} \mathrm{question}, \mathrm{your} \mathrm{should} \mathrm{have} \mathrm{complete} \mathrm{theoretical} \mathrm{knowledge} \\ \mathrm{about} \mathrm{tangent} \mathrm{and} \mathrm{normals} \mathrm{of} \mathrm{parabola} \mathrm{because} \mathrm{various} \mathrm{results} \mathrm{will} \mathrm{be} \mathrm{used} \\ \mathrm{directly}. \\ \mathrm{We} \mathrm{have} {\mathrm{y}}^2=4\mathrm{ax} \\ \mathrm{Let} \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{P} \mathrm{and} \mathrm{Q} \mathrm{be} \left({{\mathrm{at}}_1}^2,-2{\mathrm{at}}_1\right) \mathrm{and} \left({{\mathrm{at}}_2}^2,-2{\mathrm{at}}_2\right) \mathrm{respectively}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{tangents} \mathrm{TP}\hspace{0.17em}\mathrm{and} \mathrm{TQ} \mathrm{is} \\ \left\{{\mathrm{at}}_1{\mathrm{t}}_2, \mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)\right\} \\ \mathrm{Now} \mathrm{equation} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{P} \mathrm{will} \mathrm{be}: \\ \mathrm{y}=-{\mathrm{t}}_1\mathrm{x}+2{\mathrm{at}}_1+{{\mathrm{at}}_1}^3 \\ \mathrm{Let} \mathrm{this} \mathrm{normal} \mathrm{pass} \mathrm{through} \mathrm{R} \left({{\mathrm{at}}_3}^2,-2{\mathrm{at}}_3\right) \mathrm{on} \mathrm{the} \mathrm{parabola}. \mathrm{Then} \mathrm{we} \mathrm{know} \mathrm{that} \\ {\mathrm{t}}_1+\frac{2}{{\mathrm{t}}_1}=-{\mathrm{t}}_3 ;\left(\mathrm{i}\right) \\ \mathrm{Similarly} \mathrm{if} \mathrm{normal} \mathrm{at} \mathrm{Q}\hspace{0.17em}\mathrm{passes} \mathrm{through} \mathrm{R}, \mathrm{then}: \\ {\mathrm{t}}_2+\frac{2}{{\mathrm{t}}_2}=-{\mathrm{t}}_3 ;\left(\mathrm{ii}\right) \\ \mathrm{By} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ {\mathrm{t}}_1+\frac{2}{{\mathrm{t}}_1}={\mathrm{t}}_2+\frac{2}{{\mathrm{t}}_2} \\ \Rightarrow \left({\mathrm{t}}_1-{\mathrm{t}}_2\right)+2\left(\frac{1}{{\mathrm{t}}_1}-\frac{1}{{\mathrm{t}}_2}\right)=0 \\ \Rightarrow \left({\mathrm{t}}_1-{\mathrm{t}}_2\right)-\frac{2}{{\mathrm{t}}_1{\mathrm{t}}_2}\left({\mathrm{t}}_1-{\mathrm{t}}_2\right)=0 \\ \Rightarrow \left({\mathrm{t}}_1-{\mathrm{t}}_2\right)\left(1-\frac{2}{{\mathrm{t}}_1{\mathrm{t}}_2}\right)=0 \\ \left({\mathrm{t}}_1-{\mathrm{t}}_2\right)\ne 0 \mathrm{as} \mathrm{P} \mathrm{and} \mathrm{Q}\hspace{0.17em}\mathrm{cannot} \mathrm{be} \mathrm{same}. \\ \Rightarrow 1-\frac{2}{{\mathrm{t}}_1{\mathrm{t}}_2}=0 \\ \Rightarrow {\mathrm{t}}_1{\mathrm{t}}_2=2 \\ \mathrm{We} \mathrm{want} \mathrm{centre} \mathrm{of} \mathrm{triangle} \mathrm{i}.\mathrm{e}. \mathrm{circumcentre}. \\ \mathrm{Let} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{passing} \mathrm{through} \mathrm{P},\mathrm{Q} \mathrm{and} \mathrm{T} \mathrm{be} \left(\mathrm{x},\mathrm{y}\right), \mathrm{hence} \mathrm{we} \mathrm{h} \\ 2\mathrm{x}=\mathrm{a}{\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)}^2+2\mathrm{a} ;\left(\mathrm{iii}\right) \\ 2\mathrm{y}=\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)\left(1-{\mathrm{t}}_1{\mathrm{t}}_2\right) \\ \Rightarrow 2\mathrm{y}=\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)\left(1-2\right) \left[\mathrm{As} {\mathrm{t}}_1{\mathrm{t}}_2=2\right] \\ \Rightarrow 2\mathrm{y}=-\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right) \\ \Rightarrow \left({\mathrm{t}}_1+{\mathrm{t}}_2\right)=-\frac{2\mathrm{y}}{\mathrm{a}} \\ \mathrm{Put} \mathrm{this} \mathrm{in} \left(\mathrm{iii}\right) \\ 2\mathrm{x}=\mathrm{a}{\left(-\frac{2\mathrm{y}}{\mathrm{a}}\right)}^2+2\mathrm{a} \\ 2\mathrm{x}-2\mathrm{a}=\mathrm{a}\left(\frac{4{\mathrm{y}}^2}{{\mathrm{a}}^2}\right) \\ \Rightarrow \frac{4{\mathrm{y}}^2}{\mathrm{a}}=2\left(\mathrm{x}-\mathrm{a}\right) \\ \Rightarrow \mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{a}\left(\mathbf{x}\mathbf{-}\mathbf{a}\right) \end{gathered}$$

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