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Question Number 205530 by cherokeesay last updated on 23/Mar/24
Answered by mr W last updated on 24/Mar/24
P(a cos θ, b sin θ) tan ϕ=−(dy/dx)=(b/(a tan θ)) r=a cos θ−r sin ϕ ⇒r sin ϕ=a cos θ−r r=b sin θ−r cos ϕ ⇒r cos ϕ=b sin θ−r (b/(a tan θ))=((a cos θ−r)/(b sin θ−r)) ⇒r=(((a^2 −b^2 ) sin θ)/(a tan θ−b)) r^2 =(a cos θ−r)^2 +(b sin θ−r)^2 a^2 cos^2 θ+b^2 sin^2 θ+r^2 −2r(a cos θ+b sin θ)=0 ⇒r=a cos θ+b sin θ−(√(ab sin 2θ)) ⇒(((a^2 −b^2 ) sin θ cos θ)/(a sin θ−b cos θ))=a cos θ+b sin θ−(√(ab sin 2θ)) ⇒θ≈1.2302, r≈3.3338
$${P}\left({a}\:\mathrm{cos}\:\theta,\:{b}\:\mathrm{sin}\:\theta\right) \\ $$$$\mathrm{tan}\:\varphi=−\frac{{dy}}{{dx}}=\frac{{b}}{{a}\:\mathrm{tan}\:\theta} \\ $$$${r}={a}\:\mathrm{cos}\:\theta−{r}\:\mathrm{sin}\:\varphi\:\Rightarrow{r}\:\mathrm{sin}\:\varphi={a}\:\mathrm{cos}\:\theta−{r} \\ $$$${r}={b}\:\mathrm{sin}\:\theta−{r}\:\mathrm{cos}\:\varphi\:\Rightarrow{r}\:\mathrm{cos}\:\varphi={b}\:\mathrm{sin}\:\theta−{r} \\ $$$$\frac{{b}}{{a}\:\mathrm{tan}\:\theta}=\frac{{a}\:\mathrm{cos}\:\theta−{r}}{{b}\:\mathrm{sin}\:\theta−{r}} \\ $$$$\Rightarrow{r}=\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\:\mathrm{sin}\:\theta}{{a}\:\mathrm{tan}\:\theta−{b}} \\ $$$${r}^{\mathrm{2}} =\left({a}\:\mathrm{cos}\:\theta−{r}\right)^{\mathrm{2}} +\left({b}\:\mathrm{sin}\:\theta−{r}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta+{b}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta+{r}^{\mathrm{2}} −\mathrm{2}{r}\left({a}\:\mathrm{cos}\:\theta+{b}\:\mathrm{sin}\:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow{r}={a}\:\mathrm{cos}\:\theta+{b}\:\mathrm{sin}\:\theta−\sqrt{{ab}\:\mathrm{sin}\:\mathrm{2}\theta} \\ $$$$\Rightarrow\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{{a}\:\mathrm{sin}\:\theta−{b}\:\mathrm{cos}\:\theta}={a}\:\mathrm{cos}\:\theta+{b}\:\mathrm{sin}\:\theta−\sqrt{{ab}\:\mathrm{sin}\:\mathrm{2}\theta} \\ $$$$\Rightarrow\theta\approx\mathrm{1}.\mathrm{2302},\:{r}\approx\mathrm{3}.\mathrm{3338} \\ $$
Commented by mr W last updated on 23/Mar/24
Commented by cherokeesay last updated on 23/Mar/24
thank you master !
$${thank}\:{you}\:{master}\:! \\ $$

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