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Find-eccentricity-of-the-ellipse-7x-2-7y-2-2xy-10x-10y-7-0-

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Question Number 29536 by math solver last updated on 09/Feb/18
Find eccentricity of the ellipse 7x^2 +7y^2 +2xy+10x−10y−7=0 ?
$${Find}\:{eccentricity}\:{of}\:{the}\:{ellipse} \\ $$$$\mathrm{7}{x}^{\mathrm{2}} +\mathrm{7}{y}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{10}{x}−\mathrm{10}{y}−\mathrm{7}=\mathrm{0}\:? \\ $$
Commented by math solver last updated on 10/Feb/18
sir , i did this q. as : i rotated it by 45^o and then after converting to general eq. of ellipse i got ((√3)/2). is my method is wrong or i did calculation mistake.
$${sir}\:,\:{i}\:{did}\:{this}\:{q}.\:{as}\:: \\ $$$${i}\:{rotated}\:{it}\:{by}\:\mathrm{45}^{{o}} \:{and}\:{then}\:{after} \\ $$$${converting}\:{to}\:{general}\:{eq}.\:{of}\:{ellipse} \\ $$$${i}\:{got}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\:{is}\:{my}\:{method}\:{is}\:{wrong} \\ $$$${or}\:{i}\:{did}\:{calculation}\:{mistake}. \\ $$
Commented by ajfour last updated on 10/Feb/18
it is to be rotated about its centre which is not origin, make sure!
$${it}\:{is}\:{to}\:{be}\:{rotated}\:{about}\:{its}\:{centre} \\ $$$${which}\:{is}\:{not}\:{origin},\:{make}\:{sure}! \\ $$
Commented by ajfour last updated on 10/Feb/18
Commented by ajfour last updated on 10/Feb/18
u = h+rcos (𝛗−𝛉) v = k+rsin (𝛗−𝛉) rcos 𝛗 = x ; rsin 𝛗 = y , So ⇒ u−h=xcos 𝛉+ysin 𝛉 v−k= ycos 𝛉−xsin 𝛉 Let an ellipse with centre (h,k) has eq. (((u−h)^2 )/a^2 )+(((v−k)^2 )/b^2 ) =1 After rotation by angle 𝛉, eq. is (((xcos 𝛉+ysin 𝛉)^2 )/a^2 )+(((ycos 𝛉−xsin 𝛉)^2 )/b^2 )=1 (((cos^2 θ)/a^2 )+((sin^2 θ)/b^2 ))x^2 +(((sin^2 θ)/a^2 )+((cos^2 θ)/b^2 ))y^2 + (2sin θ cos θ)((1/a^2 )−(1/b^2 ))xy+...=1 here in your question coeff. of x^2 = coeff. of y^2 ⇒ cos^2 θ =sin^2 θ =(1/2) ((coeff. of xy)/(coeff. of x^2 or y^2 )) =(2/7) ⇒ (((2sin θ cos θ)((1/a^2 )−(1/b^2 )))/(sin^2 θ((1/a^2 )+(1/b^2 )))) =(2/7) with tan θ =1 , we have ((((1/a^2 )−(1/b^2 )))/((1/2)((1/a^2 )+(1/b^2 )))) = (2/7) ⇒ (6/a^2 )=(8/b^2 ) or (b^2 /a^2 ) = (4/3) > 1 So e^2 =1−(a^2 /b^2 ) = 1−(3/4) =(1/4) Hence e =(1/2) .
$$\boldsymbol{{u}}\:=\:\boldsymbol{{h}}+\boldsymbol{{r}}\mathrm{cos}\:\left(\boldsymbol{\phi}−\boldsymbol{\theta}\right) \\ $$$$\boldsymbol{{v}}\:=\:\boldsymbol{{k}}+\boldsymbol{{r}}\mathrm{sin}\:\left(\boldsymbol{\phi}−\boldsymbol{\theta}\right) \\ $$$$\boldsymbol{{r}}\mathrm{cos}\:\boldsymbol{\phi}\:=\:\boldsymbol{{x}}\:\:\:;\:\:\boldsymbol{{r}}\mathrm{sin}\:\boldsymbol{\phi}\:=\:\boldsymbol{{y}}\:,\:{So} \\ $$$$\Rightarrow\:\:\boldsymbol{{u}}−\boldsymbol{{h}}=\boldsymbol{{x}}\mathrm{cos}\:\boldsymbol{\theta}+\boldsymbol{{y}}\mathrm{sin}\:\boldsymbol{\theta} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{v}}−\boldsymbol{{k}}=\:\boldsymbol{{y}}\mathrm{cos}\:\boldsymbol{\theta}−\boldsymbol{{x}}\mathrm{sin}\:\boldsymbol{\theta} \\ $$$${Let}\:{an}\:{ellipse}\:{with}\:{centre}\:\left(\boldsymbol{{h}},\boldsymbol{{k}}\right) \\ $$$${has}\:{eq}.\:\:\:\:\frac{\left(\boldsymbol{{u}}−\boldsymbol{{h}}\right)^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} }+\frac{\left(\boldsymbol{{v}}−\boldsymbol{{k}}\right)^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} }\:=\mathrm{1} \\ $$$$\boldsymbol{{A}}{fter}\:{rotation}\:{by}\:{angle}\:\boldsymbol{\theta},\:{eq}.\:{is} \\ $$$$\:\frac{\left(\boldsymbol{{x}}\mathrm{cos}\:\boldsymbol{\theta}+\boldsymbol{{y}}\mathrm{sin}\:\boldsymbol{\theta}\right)^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} }+\frac{\left(\boldsymbol{{y}}\mathrm{cos}\:\boldsymbol{\theta}−\boldsymbol{{x}}\mathrm{sin}\:\boldsymbol{\theta}\right)^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\:\:\left(\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{{b}^{\mathrm{2}} }\right){x}^{\mathrm{2}} +\left(\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{{b}^{\mathrm{2}} }\right){y}^{\mathrm{2}} + \\ $$$$\left(\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta\right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right){xy}+…=\mathrm{1} \\ $$$${here}\:{in}\:{your}\:{question} \\ $$$${coeff}.\:{of}\:{x}^{\mathrm{2}} =\:{coeff}.\:{of}\:{y}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\mathrm{cos}\:^{\mathrm{2}} \theta\:=\mathrm{sin}\:^{\mathrm{2}} \theta\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{coeff}.\:{of}\:{xy}}{{coeff}.\:{of}\:{x}^{\mathrm{2}} \:{or}\:{y}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\mathrm{7}} \\ $$$$\Rightarrow\:\:\:\frac{\left(\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta\right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)}{\mathrm{sin}\:^{\mathrm{2}} \theta\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)}\:=\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${with}\:\:\mathrm{tan}\:\theta\:=\mathrm{1}\:,\:{we}\:{have} \\ $$$$\:\:\:\:\:\:\frac{\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)}{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)}\:=\:\frac{\mathrm{2}}{\mathrm{7}} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\frac{\mathrm{6}}{{a}^{\mathrm{2}} }=\frac{\mathrm{8}}{{b}^{\mathrm{2}} }\:\:\:{or}\:\:\:\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:=\:\frac{\mathrm{4}}{\mathrm{3}}\:>\:\mathrm{1} \\ $$$${So}\:\:\:\:{e}^{\mathrm{2}} \:=\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${Hence}\:\:\:\boldsymbol{{e}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$
Commented by math solver last updated on 10/Feb/18
thank you sir.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by ajfour last updated on 09/Feb/18
let eq. of ellipse be (((y−mx+c_1 )^2 )/a^2 )+(((x+my+c_2 )^2 )/b^2 )=1 ⇒ ((1/a^2 )+(m^2 /b^2 ))y^2 +((1/b^2 )+(m^2 /a^2 ))x^2 + 2m((1/b^2 )−(1/a^2 ))xy+......=1 as coeff. of x^2 =coeff. of y^2 , (1/a^2 )+(m^2 /b^2 )=(1/b^2 )+(m^2 /a^2 ) as a ≠ b ⇒ m^2 =1 ((coeff. of x^2 )/(coeff. of xy)) =(7/2) ⇒ (((1/b^2 )+(m^2 /a^2 ))/(2m((1/b^2 )−(1/a^2 )))) =(7/2) or ((1+((m^2 b^2 )/a^2 ))/(m(1−(b^2 /a^2 ))))=7 ....(i) e^2 =1−(b^2 /a^2 ) =1−k^2 (say) then with m=−1, (i) becomes 1+k^2 =7k^2 −7 ⇒ k^2 =(4/3) ⇒ e^2 =−(1/3) (not possible) but with m=1 , (i) becomes 1+k^2 = 7−7k^2 ⇒ k^2 =(3/4) ⇒ e^2 =1−(3/4) or e=(1/2) .
$${let}\:{eq}.\:{of}\:{ellipse}\:{be} \\ $$$$\frac{\left({y}−{mx}+{c}_{\mathrm{1}} \right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({x}+{my}+{c}_{\mathrm{2}} \right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\:\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{{m}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right){y}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right){x}^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{m}\left(\frac{\mathrm{1}}{{b}^{\mathrm{2}} }−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right){xy}+……=\mathrm{1} \\ $$$${as}\:{coeff}.\:{of}\:{x}^{\mathrm{2}} ={coeff}.\:{of}\:{y}^{\mathrm{2}} \:,\: \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{{m}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$${as}\:\:{a}\:\neq\:{b}\:\:\:\Rightarrow\:\:\:{m}^{\mathrm{2}} =\mathrm{1} \\ $$$$\frac{{coeff}.\:{of}\:{x}^{\mathrm{2}} }{{coeff}.\:{of}\:{xy}}\:=\frac{\mathrm{7}}{\mathrm{2}}\:\:\:\Rightarrow \\ $$$$\:\:\:\:\frac{\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}{\mathrm{2}{m}\left(\frac{\mathrm{1}}{{b}^{\mathrm{2}} }−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)}\:=\frac{\mathrm{7}}{\mathrm{2}}\:\:\:\:\:\: \\ $$$${or}\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}+\frac{{m}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}{{m}\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)}=\mathrm{7}\:\:\:\:\:\:\:\:….\left({i}\right) \\ $$$${e}^{\mathrm{2}} =\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:\:\:=\mathrm{1}−{k}^{\mathrm{2}} \:\:\:\left({say}\right) \\ $$$${then}\:{with}\:{m}=−\mathrm{1},\:\:\left({i}\right)\:{becomes} \\ $$$$\mathrm{1}+{k}^{\mathrm{2}} =\mathrm{7}{k}^{\mathrm{2}} −\mathrm{7}\:\:\:\Rightarrow\:\:{k}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow\:\:\:{e}^{\mathrm{2}} \:=−\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\left({not}\:{possible}\right) \\ $$$${but}\:\:{with}\:{m}=\mathrm{1}\:,\:\:\left({i}\right)\:{becomes} \\ $$$$\mathrm{1}+{k}^{\mathrm{2}} =\:\mathrm{7}−\mathrm{7}{k}^{\mathrm{2}} \:\:\:\:\Rightarrow\:\:\:{k}^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\:{e}^{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\:\:\:{or}\:\:{e}=\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$

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