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Question Number 183728 by mr W last updated on 29/Dec/22

Commented by mr W last updated on 29/Dec/22

$$findthesmallestandthelargest$$$$equilateraltrianglewhichhastwo$$$$ofitsverticesontheparabolay=2x^2$$$$andthethirdvertexonthecircle$$$$x^2+y^2-4x+3=0.$$

Answered by mr W last updated on 29/Dec/22

Commented by mr W last updated on 29/Dec/22

$$parabola:y=2x^2$$$$circle:{(x-2)}^2+y^2=1^2$$$$letm=\tan \theta$$$$M(h,k)$$$$eqn.ofPQ:$$$$y=m(x-h)+k$$$$intersectionPandQ:$$$$2x^2=m(x-h)+k$$$$x^2-\frac{m}{2}x+\frac{mh-k}{2}=0$$$$\Rightarrow x_1+x_2=\frac{m}{2},x_1{x}_2=\frac{mh-k}{2}$$$$h=\frac{x_1+x_2}{2}=\frac{m}{4}$$$${(x_2-x_1)}^2+{(y_2-y_1)}^2=s^2$$$${(x_2+x_1)}^2-4x_1{x}_2+4{(x_2^2-x_1^2)}^2=s^2$$$${(x_2+x_1)}^2-4x_1{x}_2+4{[{(x_1+x_2)}^2-2x_1{x}_2]}^2-4{\left(2x_1{x}_2\right)}^2=s^2$$$$4h^2-2mh+2k+4{(4h^2-mh+k)}^2-4{(mh-k)}^2=s^2$$$$-4h^2+2k+4k^2-4{(4h^2-k)}^2=s^2$$$$\Rightarrow k=\frac{s^2+64h^4+4h^2}{2(16h^2+1)}=\frac{4s^2+m^4+m^2}{8(m^2+1)}=\frac{s^2}{2(m^2+1)}+\frac{m^2}{8}$$$$x_R=h+\frac{\sqrt{3}s}{2}\sin \theta =\frac{m}{4}+\frac{m\sqrt{3}s}{2\sqrt{m^2+1}}$$$$y_R=k-\frac{\sqrt{3}s}{2}\cos \theta =k-\frac{\sqrt{3}s}{2\sqrt{m^2+1}}$$$$\Rightarrow {(\frac{m}{4}+\frac{m\sqrt{3}s}{2\sqrt{m^2+1}}-2)}^2+{(\frac{s^2}{2(m^2+1)}+\frac{m^2}{8}-\frac{\sqrt{3}s}{2\sqrt{m^2+1}})}^2=1$$$$\Rightarrow s_{min}\approx 0.6771atm\approx 2.237$$$$\Rightarrow s_{max}\approx 3.5303atm\approx 1.288$$

Commented by mr W last updated on 29/Dec/22

Commented by mr W last updated on 29/Dec/22

Commented by mr W last updated on 29/Dec/22

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