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Question Number 51921 by peter frank last updated on 01/Jan/19

$$Anormalchordtoan$$$$ellipse\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$makeanangleof{45}^{°}$$$$withtheaxis.prove$$$$thatthesquareofits$$$$lengthisequalto$$$$\frac{32a^4{b}^4}{{(a^2+b^2)}^3}$$

Answered by ajfour last updated on 01/Jan/19

$$let{x}_P=a\cos \theta ,y_P=b\sin \theta$$$$-\frac{dx}{dy}{\mid }_P=\frac{a\sin \theta }{b\cos \theta }=\frac{a}{b}\tan \theta =\tan 45°=1$$$$\Rightarrow \tan \theta =\frac{b}{a}$$$$\Rightarrow a\cos \theta =\frac{a^2}{\sqrt{a^2+b^2}}=\frac{a^2}{s}\left(say\right)$$$$b\sin \theta =\frac{b^2}{s}$$$$eq.ofnormal$$$$letthisnormalintersectsellipse$$$$aottheotherpointQ(h,k)$$$$letPQ=l$$$$h=a\cos \theta -\frac{l}{\sqrt{2}},k=b\cos \theta -\frac{l}{\sqrt{2}}$$$$orQ=(\frac{a^2}{s}-\frac{l}{\sqrt{2}},\frac{b^2}{s}-\frac{l}{\sqrt{2}})$$$$asQisonellipse,$$$$b^2{(\frac{a^2}{s}-\frac{l}{\sqrt{2}})}^2+a^2{(\frac{b^2}{s}-\frac{l}{\sqrt{2}})}^2=a^2{b}^2$$$$\Rightarrow b^2{(a^2\sqrt{2}-sl)}^2+a^2{(b^2\sqrt{2}-sl)}^2=2a^2{b}^2{s}^2$$$$\Rightarrow 2a^2{b}^2(a^2+b^2)+s^2{l}^2(a^2+b^2)$$$$-2\sqrt{2}{sa}^2{b}^{2}l-2a^2{b}^2{s}^2=0$$$$\Rightarrow since{s}^2=a^2+b^2,weget$$$$l^2-\frac{2\sqrt{2}{a}^2{b}^2}{s^3}l=0$$$$\Rightarrow l^2=\frac{8a^4{b}^4}{{(a^2+b^2)}^3}.$$

Commented by peter frank last updated on 01/Jan/19

$$thanksir$$$$$$

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