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Question-44801




Question Number 44801 by ajfour last updated on 04/Oct/18
Commented by ajfour last updated on 04/Oct/18
Equation of ellipse:  (x^2 /a^2 )+(y^2 /b^2 )=1 ,  find radius r of circle.
$${Equation}\:{of}\:{ellipse}:\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:, \\ $$$${find}\:{radius}\:\boldsymbol{{r}}\:{of}\:{circle}. \\ $$
Answered by ajfour last updated on 05/Oct/18
let centre of circle be (h,0)     h = (r/(sin tan^(−1) m)) = ((r(√(1+m^2 )))/m)    point of contact (upper) : (x_0 ,y_0 )      { ((x_0 =acos φ = h+rcos θ )),((y_0 = bsin φ = rsin θ)) :}        tan φ = ((arsin θ)/(b(h+rcos θ)))   ....(i)     (dy/dx)∣_x_0  = ((bcos φ)/(−asin φ)) = −((cos θ)/(sin θ))                ⇒  tan φ = ((bsin θ)/(acos θ))   ...(ii)  comparing (i)&(ii)         ((arsin θ)/(b(h+rcos θ))) = ((bsin θ)/(acos θ))  as θ≠0 ,  a^2 rcos θ = b^2 h+b^2 rcos θ  ________________________    { ((if  θ = 0 ⇒  a = b, then)),((  r = a−h = a−((r(√(1+m^2 )))/m))) :}  ⇒   r = (a/(1+((√(1+m^2 ))/m)))    ________________________    But when its not so  ⇒      cos θ = ((b^2 h)/(r(a^2 −b^2 )))  from first bracket eqs.        (((h+rcos θ)^2 )/a^2 ) +(((rsin θ)^2 )/b^2 ) = 1  ⇒  b^2 (h+rcos θ)^2 +a^2 (r^2 −r^2 cos^2 θ)                                  = a^2 b^2   b^2 (h+((b^2 h)/(a^2 −b^2 )))^2 +a^2 [r^2 −(((b^2 h)/(a^2 −b^2 )))^2 ]                              = a^2 b^2   ⇒ r^2  = ((b^4 h^2 )/((a^2 −b^2 )^2 ))+b^2 −((a^2 b^2 h^2 )/((a^2 −b^2 )^2 ))                r^2  = b^2 (1−(h^2 /(a^2 −b^2 )))  but   h^2  = r^2 +(r^2 /m^2 )  ⇒  r^2  = b^2 −((b^2 r^2 )/(a^2 −b^2 ))−((b^2 r^2 )/(m^2 (a^2 −b^2 )))   r^2 [1+((b^2 /(a^2 −b^2 )))(1+(1/m^2 ))]= b^2   ________________________      r = (b/( (√(1+((b^2 /(a^2 −b^2 )))(1+(1/m^2 ))))))          If θ ≠ 0  (same as a ≠ b).
$${let}\:{centre}\:{of}\:{circle}\:{be}\:\left({h},\mathrm{0}\right) \\ $$$$\:\:\:\boldsymbol{{h}}\:=\:\frac{\boldsymbol{{r}}}{\mathrm{sin}\:\mathrm{tan}^{−\mathrm{1}} \boldsymbol{{m}}}\:=\:\frac{{r}\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}{{m}} \\ $$$$\:\:{point}\:{of}\:{contact}\:\left({upper}\right)\::\:\left({x}_{\mathrm{0}} ,{y}_{\mathrm{0}} \right) \\ $$$$\:\:\:\begin{cases}{{x}_{\mathrm{0}} ={a}\mathrm{cos}\:\phi\:=\:{h}+{r}\mathrm{cos}\:\theta\:}\\{{y}_{\mathrm{0}} =\:{b}\mathrm{sin}\:\phi\:=\:{r}\mathrm{sin}\:\theta}\end{cases} \\ $$$$\:\:\:\:\:\:\mathrm{tan}\:\phi\:=\:\frac{{ar}\mathrm{sin}\:\theta}{{b}\left({h}+{r}\mathrm{cos}\:\theta\right)}\:\:\:….\left({i}\right) \\ $$$$\:\:\:\frac{{dy}}{{dx}}\mid_{{x}_{\mathrm{0}} } =\:\frac{{b}\mathrm{cos}\:\phi}{−{a}\mathrm{sin}\:\phi}\:=\:−\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\mathrm{tan}\:\phi\:=\:\frac{{b}\mathrm{sin}\:\theta}{{a}\mathrm{cos}\:\theta}\:\:\:…\left({ii}\right) \\ $$$${comparing}\:\left({i}\right)\&\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:\frac{{ar}\mathrm{sin}\:\theta}{{b}\left({h}+{r}\mathrm{cos}\:\theta\right)}\:=\:\frac{{b}\mathrm{sin}\:\theta}{{a}\mathrm{cos}\:\theta} \\ $$$${as}\:\theta\neq\mathrm{0}\:,\:\:{a}^{\mathrm{2}} {r}\mathrm{cos}\:\theta\:=\:{b}^{\mathrm{2}} {h}+{b}^{\mathrm{2}} {r}\mathrm{cos}\:\theta \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\begin{cases}{{if}\:\:\theta\:=\:\mathrm{0}\:\Rightarrow\:\:\boldsymbol{{a}}\:=\:\boldsymbol{{b}},\:{then}}\\{\:\:\boldsymbol{{r}}\:=\:\boldsymbol{{a}}−\boldsymbol{{h}}\:=\:\boldsymbol{{a}}−\frac{\boldsymbol{{r}}\sqrt{\mathrm{1}+\boldsymbol{{m}}^{\mathrm{2}} }}{\boldsymbol{{m}}}}\end{cases} \\ $$$$\Rightarrow\:\:\:{r}\:=\:\frac{{a}}{\mathrm{1}+\frac{\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}{{m}}}\:\: \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:{But}\:{when}\:{its}\:{not}\:{so} \\ $$$$\Rightarrow\:\:\:\:\:\:\mathrm{cos}\:\theta\:=\:\frac{{b}^{\mathrm{2}} {h}}{{r}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)} \\ $$$${from}\:{first}\:{bracket}\:{eqs}. \\ $$$$\:\:\:\:\:\:\frac{\left({h}+{r}\mathrm{cos}\:\theta\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\frac{\left({r}\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\:{b}^{\mathrm{2}} \left({h}+{r}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \left({r}^{\mathrm{2}} −{r}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} \left({h}+\frac{{b}^{\mathrm{2}} {h}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \left[{r}^{\mathrm{2}} −\left(\frac{{b}^{\mathrm{2}} {h}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)^{\mathrm{2}} \right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Rightarrow\:{r}^{\mathrm{2}} \:=\:\frac{{b}^{\mathrm{4}} {h}^{\mathrm{2}} }{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} }+{b}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {h}^{\mathrm{2}} }{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\boldsymbol{{r}}^{\mathrm{2}} \:=\:\boldsymbol{{b}}^{\mathrm{2}} \left(\mathrm{1}−\frac{\boldsymbol{{h}}^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} }\right) \\ $$$${but}\:\:\:{h}^{\mathrm{2}} \:=\:{r}^{\mathrm{2}} +\frac{{r}^{\mathrm{2}} }{{m}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:{r}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} {r}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }−\frac{{b}^{\mathrm{2}} {r}^{\mathrm{2}} }{{m}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)} \\ $$$$\:{r}^{\mathrm{2}} \left[\mathrm{1}+\left(\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }\right)\right]=\:{b}^{\mathrm{2}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\:\:\boldsymbol{{r}}\:=\:\frac{\boldsymbol{{b}}}{\:\sqrt{\mathrm{1}+\left(\frac{\boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{m}}^{\mathrm{2}} }\right)}}\: \\ $$$$\:\:\:\:\:\:\:{If}\:\theta\:\neq\:\mathrm{0}\:\:\left({same}\:{as}\:{a}\:\neq\:{b}\right). \\ $$
Commented by ajfour last updated on 05/Oct/18
If a=5 , b=4, m=1         r = (4/( (√(1+((32)/9))) ))  = ((12)/( (√(41)))) .
$${If}\:{a}=\mathrm{5}\:,\:{b}=\mathrm{4},\:{m}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{r}}\:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{32}}{\mathrm{9}}}\:}\:\:=\:\frac{\mathrm{12}}{\:\sqrt{\mathrm{41}}}\:. \\ $$
Answered by MrW3 last updated on 05/Oct/18
eqn. of ellipse:  (x^2 /a^2 )+(y^2 /b^2 )=1  ((2x)/a^2 )+((2y)/b^2 )y′=0  ⇒y′=−((b^2 x)/(a^2 y))  center of circle (x_0 ,0) with  x_0 =((√(1+m^2 ))/m)r=r(√(1+(1/m^2 )))=μr  contact point (u,v)  u=x_0 +r cos θ=r(μ+cos θ)  v=r sin θ  ((r^2 (μ+cos θ)^2 )/a^2 )+((r^2  sin^2  θ)/b^2 )=1  (1/(tan θ))=((b^2 u)/(a^2 v))=((b^2 r(μ+cos θ))/(a^2 r sin θ))=((b^2 (μ+cos θ))/(a^2  sin θ))  a^2  cos θ=b^2 (μ+cos θ)  (a^2 −b^2 )cos θ=μb^2   cos θ=((μb^2 )/(a^2 −b^2 ))=(μ/(λ^2 −1)) with λ=(a/b)  r^2 [(((μ+cos θ)^2 )/a^2 )+((sin^2  θ)/b^2 )]=1  r^2 [((b^2 (μ+cos θ)^2 +a^2 sin^2  θ)/(a^2 b^2 ))]=1  r^2 [(((μ+(μ/(λ^2 −1)))^2 +λ^2 (1−(μ/(λ^2 −1)))(1+(μ/(λ^2 −1))))/a^2 )]=1  r^2 [(((μ^2 λ^2 (λ^2 −1)+λ^2 (λ^2 −1)^2 )/((λ^2 −1)^2 ))/a^2 )]=1  r^2 [((λ^2 (μ^2 +λ^2 −1))/(a^2 (λ^2 −1)))]=1  ⇒r=(a/λ)(√((λ^2 −1)/(μ^2 +λ^2 −1)))  ⇒r=b(√((λ^2 −1)/(μ^2 +λ^2 −1)))  ⇒r=b(√((((a/b))^2 −1)/(1+(1/m^2 )+((a/b))^2 −1)))  ⇒r=b(√((((a/b))^2 −1)/((1/m^2 )+((a/b))^2 )))  r=(b/( (√(1+(1+(1/m^2 ))((b^2 /(a^2 −b^2 )))))))
$${eqn}.\:{of}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\mathrm{2}{x}}{{a}^{\mathrm{2}} }+\frac{\mathrm{2}{y}}{{b}^{\mathrm{2}} }{y}'=\mathrm{0} \\ $$$$\Rightarrow{y}'=−\frac{{b}^{\mathrm{2}} {x}}{{a}^{\mathrm{2}} {y}} \\ $$$${center}\:{of}\:{circle}\:\left({x}_{\mathrm{0}} ,\mathrm{0}\right)\:{with} \\ $$$${x}_{\mathrm{0}} =\frac{\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}{{m}}{r}={r}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }}=\mu{r} \\ $$$${contact}\:{point}\:\left({u},{v}\right) \\ $$$${u}={x}_{\mathrm{0}} +{r}\:\mathrm{cos}\:\theta={r}\left(\mu+\mathrm{cos}\:\theta\right) \\ $$$${v}={r}\:\mathrm{sin}\:\theta \\ $$$$\frac{{r}^{\mathrm{2}} \left(\mu+\mathrm{cos}\:\theta\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{r}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\theta}=\frac{{b}^{\mathrm{2}} {u}}{{a}^{\mathrm{2}} {v}}=\frac{{b}^{\mathrm{2}} {r}\left(\mu+\mathrm{cos}\:\theta\right)}{{a}^{\mathrm{2}} {r}\:\mathrm{sin}\:\theta}=\frac{{b}^{\mathrm{2}} \left(\mu+\mathrm{cos}\:\theta\right)}{{a}^{\mathrm{2}} \:\mathrm{sin}\:\theta} \\ $$$${a}^{\mathrm{2}} \:\mathrm{cos}\:\theta={b}^{\mathrm{2}} \left(\mu+\mathrm{cos}\:\theta\right) \\ $$$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{cos}\:\theta=\mu{b}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\theta=\frac{\mu{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\frac{\mu}{\lambda^{\mathrm{2}} −\mathrm{1}}\:{with}\:\lambda=\frac{{a}}{{b}} \\ $$$${r}^{\mathrm{2}} \left[\frac{\left(\mu+\mathrm{cos}\:\theta\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{{b}^{\mathrm{2}} }\right]=\mathrm{1} \\ $$$${r}^{\mathrm{2}} \left[\frac{{b}^{\mathrm{2}} \left(\mu+\mathrm{cos}\:\theta\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\right]=\mathrm{1} \\ $$$${r}^{\mathrm{2}} \left[\frac{\left(\mu+\frac{\mu}{\lambda^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} \left(\mathrm{1}−\frac{\mu}{\lambda^{\mathrm{2}} −\mathrm{1}}\right)\left(\mathrm{1}+\frac{\mu}{\lambda^{\mathrm{2}} −\mathrm{1}}\right)}{{a}^{\mathrm{2}} }\right]=\mathrm{1} \\ $$$${r}^{\mathrm{2}} \left[\frac{\frac{\mu^{\mathrm{2}} \lambda^{\mathrm{2}} \left(\lambda^{\mathrm{2}} −\mathrm{1}\right)+\lambda^{\mathrm{2}} \left(\lambda^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{\left(\lambda^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }}{{a}^{\mathrm{2}} }\right]=\mathrm{1} \\ $$$${r}^{\mathrm{2}} \left[\frac{\lambda^{\mathrm{2}} \left(\mu^{\mathrm{2}} +\lambda^{\mathrm{2}} −\mathrm{1}\right)}{{a}^{\mathrm{2}} \left(\lambda^{\mathrm{2}} −\mathrm{1}\right)}\right]=\mathrm{1} \\ $$$$\Rightarrow{r}=\frac{{a}}{\lambda}\sqrt{\frac{\lambda^{\mathrm{2}} −\mathrm{1}}{\mu^{\mathrm{2}} +\lambda^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\Rightarrow{r}={b}\sqrt{\frac{\lambda^{\mathrm{2}} −\mathrm{1}}{\mu^{\mathrm{2}} +\lambda^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\Rightarrow{r}={b}\sqrt{\frac{\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }+\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\Rightarrow{r}={b}\sqrt{\frac{\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{1}}{\frac{\mathrm{1}}{{m}^{\mathrm{2}} }+\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} }} \\ $$$${r}=\frac{{b}}{\:\sqrt{\mathrm{1}+\left(\mathrm{1}+\frac{\mathrm{1}}{{m}^{\mathrm{2}} }\right)\left(\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)}} \\ $$
Commented by ajfour last updated on 05/Oct/18
 Thank you Sir, i hope my  answer matches..
$$\:{Thank}\:{you}\:{Sir},\:{i}\:{hope}\:{my} \\ $$$${answer}\:{matches}.. \\ $$
Commented by MrW3 last updated on 05/Oct/18
our answers are the same.
$${our}\:{answers}\:{are}\:{the}\:{same}. \\ $$

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