Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 90252 by ajfour last updated on 22/Apr/20

Commented by ajfour last updated on 22/Apr/20

If both ellipses have the same  shape, find, b/a .Also find  circumradius in terms of a.

$${If}\:{both}\:{ellipses}\:{have}\:{the}\:{same} \\ $$$${shape},\:{find},\:{b}/{a}\:.{Also}\:{find} \\ $$$${circumradius}\:{in}\:{terms}\:{of}\:{a}. \\ $$

Answered by ajfour last updated on 22/Apr/20

let eq. of red ellipse be    (((x−h)^2 )/b^2 )+(y^2 /a^2 )=1  for x=0, y^2 =b^2   ⇒  h^2 =b^2 (1−(b^2 /a^2 ))  h+b=a  ⇒   (a−b)^2 =b^2 −(b^4 /a^2 )  let  (a/b)=μ  (μ−1)^2 =1−(1/μ^2 )    ⇒  μ^4 −2μ^3 +1=0  ⇒  (μ−1)(μ^3 −μ^2 −μ−1)=0  ⇒  if μ≠1 ,    μ ≈ 1.8393

$${let}\:{eq}.\:{of}\:{red}\:{ellipse}\:{be} \\ $$$$\:\:\frac{\left({x}−{h}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$${for}\:{x}=\mathrm{0},\:{y}^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{h}^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right) \\ $$$${h}+{b}={a} \\ $$$$\Rightarrow\:\:\:\left({a}−{b}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} −\frac{{b}^{\mathrm{4}} }{{a}^{\mathrm{2}} } \\ $$$${let}\:\:\frac{{a}}{{b}}=\mu \\ $$$$\left(\mu−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{1}}{\mu^{\mathrm{2}} } \\ $$$$\:\:\Rightarrow\:\:\mu^{\mathrm{4}} −\mathrm{2}\mu^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\:\left(\mu−\mathrm{1}\right)\left(\mu^{\mathrm{3}} −\mu^{\mathrm{2}} −\mu−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{if}\:\mu\neq\mathrm{1}\:,\: \\ $$$$\:\mu\:\approx\:\mathrm{1}.\mathrm{8393} \\ $$

Commented by mr W last updated on 22/Apr/20

a solution exists:  μ^4 −2μ^3 +1=0  (μ−1)(μ^3 −μ^2 −μ−1)=0  μ=1 ⇒no solution  μ^3 −μ^2 −μ−1=0  ⇒μ=(1/3)(1+((19−3(√(33))))^(1/3) +((19+3(√(33))))^(1/3) )=1.8393

$${a}\:{solution}\:{exists}: \\ $$$$\mu^{\mathrm{4}} −\mathrm{2}\mu^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\left(\mu−\mathrm{1}\right)\left(\mu^{\mathrm{3}} −\mu^{\mathrm{2}} −\mu−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mu=\mathrm{1}\:\Rightarrow{no}\:{solution} \\ $$$$\mu^{\mathrm{3}} −\mu^{\mathrm{2}} −\mu−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mu=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{19}−\mathrm{3}\sqrt{\mathrm{33}}}+\sqrt[{\mathrm{3}}]{\mathrm{19}+\mathrm{3}\sqrt{\mathrm{33}}}\right)=\mathrm{1}.\mathrm{8393} \\ $$

Commented by mr W last updated on 22/Apr/20

Commented by ajfour last updated on 22/Apr/20

yes sir, μ≈ 1.8393 , thanks,  can we try the second part then

$${yes}\:{sir},\:\mu\approx\:\mathrm{1}.\mathrm{8393}\:,\:{thanks}, \\ $$$${can}\:{we}\:{try}\:{the}\:{second}\:{part}\:{then} \\ $$

Answered by ajfour last updated on 22/Apr/20

let center of red ellipse be   origin.  Eq. of the ellipse  (x^2 /b^2 )+(y^2 /a^2 )=1  (dy/dx)=−(a^2 /b^2 )((x/y))  let eq. of circle be  (x−p)^2 +y^2 =r^2   for  x=−(2a−b)  , y=0  ⇒  2a−b+p=r    ....(i)  let (s,t) be point of tangency  of red ellipse and circle.  (s−p)^2 +a^2 (1−(s^2 /b^2 ))=r^2   ((a^2 /b^2 )−1)s^2 +2ps+r^2 −p^2 −a^2 =0  D=0  ⇒  p^2 =((a^2 /b^2 )−1)(r^2 −p^2 −a^2 )  and as   p=b+r−2a  ((a^2 /b^2 )−1)r^2 −p^2 ((a^2 /b^2 ))=a^2 ((a^2 /b^2 )−1)  (μ^2 −1)r^2 −μ^2 (b+r−2a)^2                         =a^2 (μ^2 −1)  r^2 −4μ^2 (a−(b/2))r+a^2 (μ^2 −1)                   +4μ^2 (a−(b/2))^2 =0  let   r=λa  λ^2 −4μ^2 (1−(1/(2μ)))λ+(μ^2 −1)                          +4μ^2 (1−(1/(2μ)))^2 =0  λ=2μ^2 (1−(1/(2μ)))−(√(4μ^2 (μ^2 −1)(1−(1/(2μ)))^2 −(μ^2 −1)))  λ=μ(2μ−1)−(√((μ^2 −1){(2μ−1)^2 −1}))      with   μ=1.8393  λ≈ 1.09074

$${let}\:{center}\:{of}\:{red}\:{ellipse}\:{be}\: \\ $$$${origin}.\:\:{Eq}.\:{of}\:{the}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{dy}}{{dx}}=−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\left(\frac{{x}}{{y}}\right) \\ $$$${let}\:{eq}.\:{of}\:{circle}\:{be} \\ $$$$\left({x}−{p}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${for}\:\:{x}=−\left(\mathrm{2}{a}−{b}\right)\:\:,\:{y}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{2}{a}−{b}+{p}={r}\:\:\:\:....\left({i}\right) \\ $$$${let}\:\left({s},{t}\right)\:{be}\:{point}\:{of}\:{tangency} \\ $$$${of}\:{red}\:{ellipse}\:{and}\:{circle}. \\ $$$$\left({s}−{p}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\mathrm{1}−\frac{{s}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)={r}^{\mathrm{2}} \\ $$$$\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right){s}^{\mathrm{2}} +\mathrm{2}{ps}+{r}^{\mathrm{2}} −{p}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${D}=\mathrm{0} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{2}} =\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right)\left({r}^{\mathrm{2}} −{p}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$${and}\:{as}\:\:\:{p}={b}+{r}−\mathrm{2}{a} \\ $$$$\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right){r}^{\mathrm{2}} −{p}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)={a}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right) \\ $$$$\left(\mu^{\mathrm{2}} −\mathrm{1}\right){r}^{\mathrm{2}} −\mu^{\mathrm{2}} \left({b}+{r}−\mathrm{2}{a}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}^{\mathrm{2}} \left(\mu^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${r}^{\mathrm{2}} −\mathrm{4}\mu^{\mathrm{2}} \left({a}−\frac{{b}}{\mathrm{2}}\right){r}+{a}^{\mathrm{2}} \left(\mu^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}\mu^{\mathrm{2}} \left({a}−\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:\:\:{r}=\lambda{a} \\ $$$$\lambda^{\mathrm{2}} −\mathrm{4}\mu^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\mu}\right)\lambda+\left(\mu^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}\mu^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\mu}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\lambda=\mathrm{2}\mu^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\mu}\right)−\sqrt{\mathrm{4}\mu^{\mathrm{2}} \left(\mu^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\mu}\right)^{\mathrm{2}} −\left(\mu^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$\lambda=\mu\left(\mathrm{2}\mu−\mathrm{1}\right)−\sqrt{\left(\mu^{\mathrm{2}} −\mathrm{1}\right)\left\{\left(\mathrm{2}\mu−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right\}} \\ $$$$\:\:\:\:{with}\:\:\:\mu=\mathrm{1}.\mathrm{8393} \\ $$$$\lambda\approx\:\mathrm{1}.\mathrm{09074} \\ $$

Commented by mr W last updated on 22/Apr/20

very nice solution sir! exact!

$${very}\:{nice}\:{solution}\:{sir}!\:{exact}! \\ $$

Commented by ajfour last updated on 22/Apr/20

Terms of Service

Privacy Policy

Contact: info@tinkutara.com