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Question Number 29728 by rahul 19 last updated on 11/Feb/18

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

$$\mathrm{i}\mathrm{dnt}\mathrm{know}\mathrm{exactly}\mathrm{but}\mathrm{i}\mathrm{think}$$$$\mathrm{tangents}\mathrm{at}\mathrm{end}\mathrm{of}\mathrm{latus}\mathrm{rectum}$$$$\mathrm{can}\mathrm{form}\mathrm{square}$$

Commented by rahul 19 last updated on 11/Feb/18

Commented by rahul 19 last updated on 11/Feb/18

$$thisisthesol.$$$$ihaveadoubthow,(a^2-7)+(13-5a)$$$$=a^2.$$

Commented by rahul 19 last updated on 11/Feb/18

$$iknowdirectoreq.is{x}^2+y^2=a^2+b^2.$$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

$$\mathrm{in}\mathrm{triangle}\mathrm{BOA}...\mathrm{BO}=\mathrm{OA}=\mathrm{a}$$$$\left(\mathrm{right}\mathrm{triangle}\right)$$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

$$\mathrm{actually}...{\mathrm{a}}^2+{\mathrm{b}}^2\mathrm{is}{\mathrm{radius}}^2..\mathrm{agree}?$$$$\mathrm{but}\mathrm{radius}\mathrm{is}\mathrm{also}\mathrm{equal}\mathrm{to}\mathrm{squares}$$$$\mathrm{half}\mathrm{diagonal}=\mathrm{BO}=\mathrm{a}$$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

$$\mathrm{it}\mathrm{is}{\mathrm{radius}}^2\mathrm{of}\mathrm{director}\mathrm{circle}\mathrm{and}$$$$\mathrm{director}\mathrm{circle}\mathrm{radius}\mathrm{also}\mathrm{equal}\mathrm{to}$$$$\mathrm{BO}...\mathrm{equate}$$

Commented by rahul 19 last updated on 11/Feb/18

$$ok,fine.$$$$thanks!!$$

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