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Question Number 112576 by ajfour last updated on 08/Sep/20

Commented by ajfour last updated on 08/Sep/20

$$FindOA=AB.$$

Answered by mr W last updated on 09/Sep/20

$$A(a,a^2)$$$$B(-b,b^2)$$$$eqn.ofOA:$$$$y=ax$$$$eqn.ofAB:$$$$y=a^2-\frac{1}{a}(x-a)=a^2+1-\frac{x}{a}$$$$b^2=a^2+1+\frac{b}{a}$$$$a^2-b^{2}a+b+1=0$$$$\Rightarrow a=\frac{b^2\pm \sqrt{b^4-4(b+1)}}{2}$$$${(b^2-a^2)}^2+{(a+b)}^2=a^4+a^2$$$$2{ba}^2-2a-b(1+b^2)=0$$$$\Rightarrow a=\frac{1+\sqrt{1+2b^2(1+b^2)}}{2b}$$$$...$$

Commented by mr W last updated on 09/Sep/20

$$youareright.thankssir!$$

Answered by ajfour last updated on 08/Sep/20

$$A(a\cos \theta ,a\sin \theta );OA=AB=a$$$$\Rightarrow a\cos {}^2\theta =\sin \theta ...\left(i\right)$$$$B(a\cos \theta -a\sin \theta ,a\sin \theta +a\cos \theta )$$$$\Rightarrow a{(\sin \theta -\cos \theta )}^2=\sin \theta +\cos \theta ...\left(ii\right)$$$$Ifm=\tan \theta \left(ii\right)/\left(i\right)gives$$$${(m-1)}^2=1+\frac{1}{m}$$$$\Rightarrow m^3-2m^2-1=0...\left(i\right)$$$$m^6=4m^4+4m^2+1$$$$asa=m\sqrt{1+m^2}\Rightarrow$$$$m^6=4a^2+1$$$$from\left(i\right)8m^6=m^9-1-3m^6+3m^3$$$$\Rightarrow {(11m^6+1)}^2=m^6{(m^6+3)}^2$$$$\Rightarrow {(44a^2+12)}^2=(4a^2+1){(4a^2+4)}^2$$$$\Rightarrow 121a^4+66a^2+9=4a^6+9a^4+6a^2+1$$$$\Rightarrow a^6-28a^4-15a^2-2=0$$$$\Rightarrow a\approx 5.34118$$

Commented by ajfour last updated on 08/Sep/20

Commented by mr W last updated on 09/Sep/20

$$nicesolution!$$

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