The-equation-of-an-ellipse-is-given-as-x-2-y-2xcot2-2-1-0-lt-lt-0-25pi-Show-that-the-minimum-and-maximum-distances-from-the-centre-to-the-circumference-of-this-ellipse-are-tan-and-cot

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Question Number 2168 by Yozzi last updated on 06/Nov/15

$$Theequationofanellipseisgiven$$$$as$$$$x^2+{(y+2xcot2\theta )}^2=1(0<\theta <0.25\pi ).$$$$Showthattheminimumandmaximum$$$$distancesfromthecentretothe$$$$circumferenceofthisellipseare$$$$tan\theta andcot\theta respectively.$$$$$$

Answered by prakash jain last updated on 07/Nov/15

$$x^2+y^2+4xy\cot 2\theta +4x^2{\cot }^{2}2\theta -1=0$$$$x^2(1+{4\cot }^{2}2\theta )+y^2+4xy\cot 2\theta -1=0$$$$x=x_1\cos \alpha +y_1\sin \alpha$$$$y=-x_1\sin \alpha +y_1\cos \alpha$$$$xyco-efficientafterrotation$$$$2(1+{4\cot }^{2}2\theta )\cos \alpha \sin \alpha -2\sin \alpha \cos \alpha +4\cos 2\alpha \cot 2\theta =0$$$${4\cot }^{2}2\theta \sin 2\alpha +4\cos 2\alpha \cot 2\theta =0$$$$\alpha =-\theta$$$$\mathrm{New}\mathrm{coefficient}\mathrm{after}\mathrm{rotation}$$$${ax}^2+bxy+{cy}^2+dx+ey+f=0$$$$a=(1+{4\cot }^{2}2\theta ){\cos }^2\theta -4\cot 2\theta \sin \theta \cos \theta +{\sin }^2\theta$$$$b=0$$$$c=(1+{4\cot }^{2}2\theta ){\sin }^2\theta +4\cot 2\theta \sin \theta \cos \theta +{\cos }^2\theta$$$$d=0$$$$e=0$$$$f=1$$$$moresimplicationtobedonebutitshould$$$$reducetotheform$$$$\frac{x_1^2}{A^2}+\frac{y_1^2}{B^2}=1$$$$$$

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