Question-45477

Question Number 45477 by ajfour last updated on 13/Oct/18

Commented by ajfour last updated on 13/Oct/18

If radius of circle is unity, find equation of ellipse. (circle may or may not touch y-axes)

$${If}\:{radius}\:{of}\:{circle}\:{is}\:{unity}, \\ $$$${find}\:{equation}\:{of}\:{ellipse}. \\ $$$$\left({circle}\:{may}\:{or}\:{may}\:{not}\:{touch}\:{y}-{axes}\right) \\ $$

Commented by ajfour last updated on 13/Oct/18

Commented by MJS last updated on 13/Oct/18

it seems not unique to me at first sight but let′s try...

$$\mathrm{it}\:\mathrm{seems}\:\mathrm{not}\:\mathrm{unique}\:\mathrm{to}\:\mathrm{me}\:\mathrm{at}\:\mathrm{first}\:\mathrm{sight}\:\mathrm{but} \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{try}… \\ $$

Commented by ajfour last updated on 14/Oct/18

let eq. of ellipse be ax^2 +by^2 +2hxy+2gx+2fy+c=0 since it has a double root with y=0 ⇒ the equation ax^2 +2gx+c= 0 has D=0 ⇒ g^2 = ac ....(i) double root with x=0 ⇒ by^2 +2fy+c = 0 has D=0 ⇒ f^( 2) = bc ...(ii) Let slope of PQ= tan θ = m { ((x_Q = 2cos^2 θ )),((y_Q = 2cos θsin θ)) :} .....(iii) if (p,q) be center of ellipse ⇒ q = pm ⇒ ((af−gh)/(h^2 −ab)) = m(((bg−hf)/(h^2 −ab))) ...(iv) OQ=r = (√(p^2 +q^2 ))−2cos θ { ((x_Q = p−rcos θ)),((y_Q = q−rsin θ)) :} ......(v) differentiating eq. of ellipse 2ax+2by(dy/dx)+2hx(dy/dx)+2hy+2g+ 2f (dy/dx) = 0 (dy/dx)∣_Q = −(1/m) ....(vi) And Q(x_Q ,y_Q ) lies on ellipse so (x_Q , y_Q ) satisfies ax^2 +by^2 +2hxy+2gx+2fy+c=0 ....(vii) ________________________

$${let}\:{eq}.\:{of}\:{ellipse}\:{be}\: \\ $$$${ax}^{\mathrm{2}} +{by}^{\mathrm{2}} +\mathrm{2}{hxy}+\mathrm{2}{gx}+\mathrm{2}{fy}+{c}=\mathrm{0} \\ $$$${since}\:{it}\:{has}\:{a}\:{double}\:{root}\:{with} \\ $$$${y}=\mathrm{0}\:\Rightarrow\:{the}\:{equation} \\ $$$$\:\:\:\:{ax}^{\mathrm{2}} +\mathrm{2}{gx}+{c}=\:\mathrm{0}\:\:{has}\:{D}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\:\boldsymbol{{g}}^{\mathrm{2}} \:=\:\boldsymbol{{ac}}\:\:\:\:\:\:….\left(\boldsymbol{{i}}\right) \\ $$$${double}\:{root}\:{with}\:{x}=\mathrm{0}\:\:\Rightarrow \\ $$$$\:\:\:\:\:{by}^{\mathrm{2}} +\mathrm{2}{fy}+{c}\:=\:\mathrm{0}\:\:{has}\:{D}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\:\:\boldsymbol{{f}}^{\:\mathrm{2}} \:=\:\boldsymbol{{bc}}\:\:\:\:\:\:\:…\left(\boldsymbol{{ii}}\right) \\ $$$${Let}\:{slope}\:{of}\:{PQ}=\:\mathrm{tan}\:\theta\:=\:{m} \\ $$$$\:\:\:\begin{cases}{{x}_{{Q}} =\:\mathrm{2cos}\:^{\mathrm{2}} \theta\:}\\{{y}_{{Q}} =\:\mathrm{2cos}\:\theta\mathrm{sin}\:\theta}\end{cases}\:\:\:\:\:\:…..\left({iii}\right) \\ $$$${if}\:\:\left({p},{q}\right)\:{be}\:{center}\:{of}\:{ellipse} \\ $$$$\Rightarrow\:\:{q}\:=\:{pm} \\ $$$$\Rightarrow\:\:\:\frac{{af}−{gh}}{{h}^{\mathrm{2}} −{ab}}\:=\:{m}\left(\frac{{bg}−{hf}}{{h}^{\mathrm{2}} −{ab}}\right)\:\:\:\:…\left({iv}\right) \\ $$$$\:{OQ}={r}\:=\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }−\mathrm{2cos}\:\theta \\ $$$$\:\:\:\:\begin{cases}{{x}_{{Q}} =\:{p}−{r}\mathrm{cos}\:\theta}\\{{y}_{{Q}} =\:{q}−{r}\mathrm{sin}\:\theta}\end{cases}\:\:\:\:\:\:……\left({v}\right) \\ $$$$\:\:{differentiating}\:{eq}.\:{of}\:{ellipse} \\ $$$$\mathrm{2}{ax}+\mathrm{2}{by}\frac{{dy}}{{dx}}+\mathrm{2}{hx}\frac{{dy}}{{dx}}+\mathrm{2}{hy}+\mathrm{2}{g}+ \\ $$$$\:\:\:\:\mathrm{2}{f}\:\frac{{dy}}{{dx}}\:=\:\mathrm{0} \\ $$$$\:\:\frac{{dy}}{{dx}}\mid_{{Q}} \:=\:−\frac{\mathrm{1}}{{m}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({vi}\right) \\ $$$${And}\:{Q}\left({x}_{{Q}} \:,{y}_{{Q}} \right)\:{lies}\:{on}\:{ellipse}\: \\ $$$${so}\:\left({x}_{{Q}} ,\:{y}_{{Q}} \right)\:{satisfies}\: \\ $$$${ax}^{\mathrm{2}} +{by}^{\mathrm{2}} +\mathrm{2}{hxy}+\mathrm{2}{gx}+\mathrm{2}{fy}+{c}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({vii}\right) \\ $$$$\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$

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