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Question Number 70738 by ajfour last updated on 07/Oct/19

Commented by ajfour last updated on 07/Oct/19

$$Find\frac{R}{a}\cdot$$

Commented by ajfour last updated on 07/Oct/19

Answered by mr W last updated on 08/Oct/19

$$C(k,R)$$$$P(-h,h^2)$$$$y^{\prime }=2x=-2h$$$$eqn.ofPC:$$$$y=\frac{x+h}{2h}+h^2$$$$R=\frac{k+h}{2h}+h^2$$$$\Rightarrow k+h=2h(R-h^2)$$$${(k+h)}^2+{(R-h^2)}^2=R^2$$$$4h^2{(R-h^2)}^2+{(R-h^2)}^2=R^2$$$$4R^2-8{Rh}^2+4h^4-2R+h^2=0$$$$4h^4-(8R-1)h^2+2R(2R-1)=0$$$$h^2=\frac{8R-1-\sqrt{16R+1}}{8}$$$$\Rightarrow h=\frac{1}{2}\sqrt{\frac{8R-1-\sqrt{16R+1}}{2}}$$$$\Rightarrow k=[2(R-h^2)-1]h$$$$$$$$centerofsmallcircleA(-r,p)$$$${(k+r)}^2+{(R-p)}^2={(R+r)}^2$$$${(R-p)}^2=(R-k)(R+k+2r)$$$$p=R-\sqrt{(R-k)(R+k+2r)}$$$$$$$$T(-s,s^2)$$$$y^{\prime }=-2s$$$$eqn.ofTA:$$$$y=\frac{x+s}{2s}+s^2$$$$p=\frac{-r+s}{2s}+s^2$$$$p-s^2=\frac{s-r}{2s}$$$${(-r+s)}^2+{(p-s^2)}^2=r^2$$$$4s^4-8{rs}^3+s^2-2rs+r^2=0$$$$r^2-2s(4s^2+1)r+s^2(4s^2+1)=0$$$$\Rightarrow r=s(4s^2+1)-2s^2\sqrt{4s^2+1}...\left(i\right)$$$$p=s^2+\frac{s-r}{2s}=R-\sqrt{(R-k)(R+k+2r)}...\left(ii\right)$$$$wegetsfrom\left(ii\right)forgivenR,$$$$andthenrfrom\left(i\right).$$$$examples:$$$$R=8\Rightarrow r=0.6351$$$$R=4\Rightarrow r=0.2956$$

Commented by mr W last updated on 08/Oct/19

Commented by mr W last updated on 08/Oct/19

Commented by ajfour last updated on 08/Oct/19

$$thanksSir.Greatattempt!$$

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