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 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) .

$${let}\:{eq}.\:{of}\:{ellipse}\:{be} \\ $$$$\frac{\left({y}−{mx}+{c}_{\mathrm{1}} \right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({x}+{my}+{c}_{\mathrm{2}} \right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\:\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{{m}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right){y}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right){x}^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{m}\left(\frac{\mathrm{1}}{{b}^{\mathrm{2}} }−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right){xy}+......=\mathrm{1} \\ $$$${as}\:{coeff}.\:{of}\:{x}^{\mathrm{2}} ={coeff}.\:{of}\:{y}^{\mathrm{2}} \:,\: \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{{m}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$${as}\:\:{a}\:\neq\:{b}\:\:\:\Rightarrow\:\:\:{m}^{\mathrm{2}} =\mathrm{1} \\ $$$$\frac{{coeff}.\:{of}\:{x}^{\mathrm{2}} }{{coeff}.\:{of}\:{xy}}\:=\frac{\mathrm{7}}{\mathrm{2}}\:\:\:\Rightarrow \\ $$$$\:\:\:\:\frac{\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}{\mathrm{2}{m}\left(\frac{\mathrm{1}}{{b}^{\mathrm{2}} }−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)}\:=\frac{\mathrm{7}}{\mathrm{2}}\:\:\:\:\:\: \\ $$$${or}\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}+\frac{{m}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}{{m}\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)}=\mathrm{7}\:\:\:\:\:\:\:\:....\left({i}\right) \\ $$$${e}^{\mathrm{2}} =\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:\:\:=\mathrm{1}−{k}^{\mathrm{2}} \:\:\:\left({say}\right) \\ $$$${then}\:{with}\:{m}=−\mathrm{1},\:\:\left({i}\right)\:{becomes} \\ $$$$\mathrm{1}+{k}^{\mathrm{2}} =\mathrm{7}{k}^{\mathrm{2}} −\mathrm{7}\:\:\:\Rightarrow\:\:{k}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow\:\:\:{e}^{\mathrm{2}} \:=−\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\left({not}\:{possible}\right) \\ $$$${but}\:\:{with}\:{m}=\mathrm{1}\:,\:\:\left({i}\right)\:{becomes} \\ $$$$\mathrm{1}+{k}^{\mathrm{2}} =\:\mathrm{7}−\mathrm{7}{k}^{\mathrm{2}} \:\:\:\:\Rightarrow\:\:\:{k}^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\:{e}^{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\:\:\:{or}\:\:{e}=\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com