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Question Number 70503 by peter frank last updated on 04/Oct/19

Commented by Prithwish sen last updated on 06/Oct/19

$$\mathrm{Let}\mathrm{O}=({\mathrm{x}}_1,{\mathrm{y}}_1)$$$$\because \mathrm{It}\mathrm{is}\mathrm{on}\mathrm{the}\mathrm{circle}$$$$\therefore {\mathbf{x}}_1^2+{\mathbf{y}}_1^2={\mathbf{d}}^2.....\left(\mathbf{i}\right)$$$$\mathrm{Now}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathbf{chord}\mathbf{of}\mathbf{contact}$$$$\mathbf{PQ}\equiv \frac{{\mathbf{xx}}_1}{{\mathbf{a}}^2}+\frac{{\mathbf{yy}}_1}{{\mathbf{b}}^2}=1....\left(\mathbf{ii}\right)$$$$\mathrm{Let}\mathrm{the}\mathrm{midpoint}\mathrm{of}\mathbf{P}({\mathbf{x}}_2,{\mathbf{y}}_2)\mathrm{and}\mathbf{Q}({\mathbf{x}}_3,{\mathbf{y}}_3)$$$$\mathrm{be}(\mathbf{h},\mathbf{k})$$$$\mathbf{i}.\mathbf{e}{\mathbf{x}}_2+{\mathbf{x}}_3=2\mathbf{h}.....\left(\mathbf{iii}\right)$$$$\mathbf{and}{\mathbf{y}}_2+{\mathbf{y}}_3=2\mathbf{k}.....\left(\mathbf{iv}\right)$$$$\mathrm{as}\left(\mathrm{ii}\right)\mathrm{intersects}\mathrm{the}\mathrm{ellipse}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1...\left(\mathrm{v}\right)$$$$\mathrm{putting}\frac{\mathrm{y}}{\mathrm{b}}=\frac{\mathrm{b}}{{\mathrm{y}}_1}(1-\frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2})\mathrm{from}\left(\mathrm{ii}\right)\mathrm{to}\left(\mathrm{v}\right)$$$$\mathrm{we}\mathrm{get},$$$$\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{b}}^2}{{\mathrm{y}}_1^2}{(1-\frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2})}^2=1$$$$\Rightarrow {\mathbf{x}}^2\{\frac{1}{{\mathbf{a}}^2}+\frac{{\mathbf{b}}^2{\mathbf{x}}_1^2}{{\mathbf{a}}^4{\mathbf{y}}_1^2}\}-\mathbf{x}.2\frac{{\mathbf{x}}_1}{{\mathbf{y}}_1^2}.\frac{{\mathbf{b}}^2}{{\mathbf{a}}^2}+(\frac{{\mathbf{b}}^2}{{\mathbf{y}}_1^2}-1)=0$$$$\therefore {\mathbf{x}}_2+{\mathbf{x}}_3=\frac{2{\mathbf{x}}_1{\mathbf{a}}^2{\mathbf{b}}^2}{({\mathbf{a}}^2{\mathbf{y}}_1^2+{\mathbf{b}}^2{\mathbf{x}}_1^2)}$$$$\mathbf{h}=\frac{{\mathbf{x}}_1{\mathbf{a}}^2{\mathbf{b}}^2}{({\mathbf{a}}^2{\mathbf{y}}_1^2+{\mathbf{b}}^2{\mathbf{x}}_1^2)}$$$$\mathbf{similarly}\mathbf{we}\mathbf{can}\mathbf{get}$$$$\mathbf{k}=\frac{{\mathbf{y}}_1{\mathbf{a}}^2{\mathbf{b}}^2}{({\mathbf{a}}^2{\mathbf{y}}_1^2+{\mathbf{b}}^2{\mathbf{x}}_1^2)}$$$$\mathbf{now}$$$${\mathbf{h}}^2+{\mathbf{k}}^2=\frac{{\mathbf{a}}^4{\mathbf{b}}^4}{{({\mathbf{a}}^2{\mathbf{y}}_1^2+{\mathbf{b}}^2{\mathbf{x}}_1^2)}^2}\{{\mathbf{x}}_1^2+{\mathbf{y}}_1^2\}$$$${\mathbf{h}}^2+{\mathbf{k}}^2={\mathbf{d}}^2.\frac{1}{{(\frac{{\mathbf{x}}_1^2}{{\mathbf{a}}^2}+\frac{{\mathbf{y}}_1^2}{{\mathbf{b}}^2})}^2}\because \mathbf{from}\left(\mathbf{i}\right)$$$$\therefore \mathbf{the}\mathbf{locus}\mathbf{of}\mathbf{the}\mathbf{mipoint}\mathbf{is}$$$${\mathbf{x}}^2+{\mathbf{y}}^2={\mathbf{d}}^2{[\frac{{\mathbf{x}}_1^2}{{\mathbf{a}}^2}+\frac{{\mathbf{y}}_1^2}{{\mathbf{b}}^2}]}^{-2}$$$$\mathbf{I}\mathbf{get}\mathbf{this}.\mathbf{Is}\mathbf{the}\mathbf{given}\mathbf{question}\mathbf{is}\mathbf{correct}?$$$$\mathbf{please}\mathbf{check}.$$

Commented by peter frank last updated on 06/Oct/19

$$thankverymuch$$

Commented by peter frank last updated on 06/Oct/19

$$thankverymuch$$

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