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Question Number 49148 by mr W last updated on 03/Dec/18

Commented by mr W last updated on 04/Dec/18

$$Thisisnotaquestion,butanuseful$$$$information.$$$$$$$$Maybesomeonehasdelighttoproveit.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Dec/18

$$thankyousir...$$

Answered by ajfour last updated on 04/Dec/18

$$\frac{{(x-u)}^2}{a^2}+\frac{{(y-v)}^2}{b^2}=1$$$$\frac{x-u}{a^2}+\frac{(y-v)}{b^2}\frac{dy}{dx}=0$$$$letpointoftangencybeP(x_1,y_1)$$$$let{x}_1=u+a\cos \theta ,y_1=v+b\sin \theta$$$$\frac{dy}{dx}{\mid }_P=-\frac{b}{a}\left(\frac{\cos \theta }{\sin \theta }\right)=-\frac{p}{q}$$$$\Rightarrow \tan \theta =\frac{bq}{ap}$$$$\Rightarrow Pliesontangent,\Rightarrow$$$$p(u+a\cos \theta )+q(v+b\sin \theta )+r=0$$$$pu+qv+r\pm \sqrt{a^2{p}^2+b^2{q}^2}=0$$$$\Rightarrow a^2{p}^2+b^2{q}^2={(pu+qv+r)}^2.$$

Commented by mr W last updated on 04/Dec/18

$$thankyoufortheproofsir!$$$$thisresultisveryhelpfulinsolving$$$$manyquestionsrelatingtoellipse.$$

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