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Question Number 103147 by ajfour last updated on 13/Jul/20

Answered by mr W last updated on 13/Jul/20

$$A(a,0)$$$$B(0,\sqrt{p^2-a^2})$$$$C(0,c)$$$$D(\sqrt{q^2-c^2},0)$$$$a\sqrt{p^2-a^2}=c\sqrt{q^2-c^2}...\left(i\right)$$$$intersectionABandCDat(h,k)$$$$\frac{h}{a}+\frac{k}{\sqrt{p^2-a^2}}=1$$$$\frac{h}{\sqrt{q^2-c^2}}+\frac{k}{c}=1$$$$(\frac{1}{ac}-\frac{1}{\sqrt{(p^2-a^2)(q^2-c^2)}})h=\frac{1}{c}-\frac{1}{\sqrt{p^2-a^2}}$$$$\Rightarrow h=\frac{\frac{1}{c}-\frac{1}{\sqrt{p^2-a^2}}}{\frac{1}{ac}-\frac{1}{\sqrt{(p^2-a^2)(q^2-c^2)}}}$$$$(c-\sqrt{p^2-a^2})h=\frac{1}{2}c\sqrt{q^2-c^2}$$$$\Rightarrow (c-\sqrt{p^2-a^2})\frac{\frac{1}{c}-\frac{1}{\sqrt{p^2-a^2}}}{\frac{1}{ac}-\frac{1}{\sqrt{(p^2-a^2)(q^2-c^2)}}}=\frac{1}{2}c\sqrt{q^2-c^2}...\left(ii\right)$$$$from\left(i\right):$$$$c^4-q^2{c}^2+a^2(p^2-a^2)=0$$$$c^2=\frac{q^2+\sqrt{q^4-4a^2(p^2-a^2)}}{2}$$$$c=\sqrt{\frac{q^2+\sqrt{q^4-4a^2(p^2-a^2)}}{2}}$$$$putthisinto\left(ii\right)tofinda...$$

Commented by mr W last updated on 13/Jul/20

Commented by ajfour last updated on 13/Jul/20

$$ThanksSir,butithink,iamable$$$$tofindtheexactexpression,$$$$kindlycheckthesame..$$

Answered by ajfour last updated on 13/Jul/20

$$let\mathrm{\angle }OAB=\theta ;\mathrm{\angle }OCD=\varphi$$$$A_1=A_2=A_3=△$$$$4△=p^2\sin \theta \cos \theta =q^2\sin \varphi \cos \varphi$$$$......\left(i\right)$$$$LetPbepointofintersectionof$$$$ABandCD.$$$$AlsosayAP=r,CP=s$$$$OD=q\sin \varphi ,OB=p\sin \theta$$$$2△=(p\cos \theta -q\sin \varphi )\left(r\sin \theta \right)..\left(ii\right)$$$$2△=(q\cos \varphi -p\sin \theta )\left(s\sin \varphi \right)..\left(iii\right)$$$$2△=\left(q\sin \varphi \right)\left(r\sin \theta \right)$$$$+\left(p\sin \theta \right)\left(s\sin \varphi \right)...\left(iv\right)$$$$using\left(ii\right),\left(iii\right)in\left(iv\right)$$$$\frac{q\sin \varphi }{p\cos \theta -q\sin \varphi }+\frac{p\sin \theta }{q\cos \varphi -p\sin \theta }=1$$$$\Rightarrow \frac{1}{\left(\frac{p\cos \theta }{q\sin \varphi }\right)-1}+\frac{1}{\left(\frac{q\cos \varphi }{p\sin \theta }\right)-1}=1$$$$$$$$Andfrom\left(i\right)$$$$\frac{q\cos \varphi }{p\sin \theta }=\frac{p\cos \theta }{q\sin \varphi }=R....\left(I\right)$$$$So\frac{1}{R-1}+\frac{1}{R-1}=1$$$$\Rightarrow R-1=2\Rightarrow R=3$$$$henceq\cos \varphi =3p\sin \theta$$$$q\sin \varphi =\sqrt{q^2-9p^2\sin {}^2\theta }$$$$substitutingthisin\left(I\right):$$$$\Rightarrow \frac{p\cos \theta }{\sqrt{q^2-9p^2\sin {}^2\theta }}=3$$$$\Rightarrow p^2\cos {}^2\theta =9q^2-81p^2(1-\cos {}^2\theta )$$$$\Rightarrow p^2\cos {}^2\theta =\frac{81p^2-9q^2}{80}$$$$\mathrm{\&}{q}^2\cos {}^2\varphi =9(p^2-p^2\cos {}^2\theta )$$$$=9(p^2-\frac{81p^2-9q^2}{80})$$$$\Rightarrow q^2\cos {}^2\varphi =\frac{81q^2-9p^2}{80}$$$${AC}^2=p^2\cos {}^2\theta +q^2\cos {}^2\varphi$$$$\Rightarrow AC{}^2=\frac{9}{10}(p^2+q^2)$$$$AC=3\sqrt{\frac{p^2+q^2}{10}}.$$$$eg.p=5,q=4$$$$AC=3\sqrt{\frac{25+16}{10}}=3\sqrt{\frac{41}{10}}$$$$p\cos \theta =\frac{3}{4}\sqrt{\frac{9p^2-q^2}{5}}$$$$=\frac{3}{4}\sqrt{\frac{225-16}{5}}=\frac{3}{4}\sqrt{\frac{209}{5}}$$$$p\sin \theta =\sqrt{25-\frac{9\times 209}{80}}$$$$=\frac{1}{4}\sqrt{\frac{119}{5}}$$$$q\cos \varphi =\frac{3}{4}\sqrt{\frac{9q^2-p^2}{5}}$$$$=\frac{3}{4}\sqrt{\frac{144-25}{5}}=\frac{3}{4}\sqrt{\frac{119}{5}}$$$$q\sin \varphi =\sqrt{16-\frac{9\times 119}{80}}=\frac{1}{4}\sqrt{\frac{209}{5}}$$$$Ishalltrytographit...$$$$$$

Commented by ajfour last updated on 13/Jul/20

Commented by mr W last updated on 13/Jul/20

$$fantasticallysolved!$$

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