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Question Number 210457 by peter frank last updated on 09/Aug/24
Commented by peter frank last updated on 09/Aug/24
(b) and (c) help
$$\left(\mathrm{b}\right)\:\mathrm{and}\:\left(\mathrm{c}\right)\:\mathrm{help} \\ $$
Commented by mr W last updated on 10/Aug/24
(b) is wrong. the locus is (x^2 /a^4 )+(y^2 /b^4 )=(1/(a^2 +b^2 ))
$$\left({b}\right)\:{is}\:{wrong}. \\ $$$${the}\:{locus}\:{is} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{4}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{4}} }=\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$
Commented by peter frank last updated on 11/Aug/24
thanks sir.please help
$$\mathrm{thanks}\:\mathrm{sir}.\mathrm{please}\:\mathrm{help} \\ $$
Answered by mr W last updated on 10/Aug/24
Commented by mr W last updated on 10/Aug/24
(c) say the parameters of the ellipse are a, b with a≥b and μ=(b/a)≤1. say P(a cos θ, b sin θ) tan α=((b tan θ)/a)=μ tan θ tan β=(b/(a tan θ))=(μ/(tan θ)) say Q(a cos ϕ, −b sin ϕ) tan β=((b tan ϕ)/a)=μ tan ϕ tan α=(b/(a tan ϕ))=(μ/(tan ϕ)) ⇒tan θ tan ϕ=1 Φ=tan (α+β) =((tan α+tan β)/(1−tan α tan β)) =((μ tan θ+μ tan ϕ)/(1−μ^2 tan θ tan ϕ)) =((μ(tan θ+(1/(tan θ))))/(1−μ^2 ))≥((2μ)/(1−μ^2 )) Φ_(min) =((2μ)/(1−μ^2 )) when tan θ=(1/(tan θ)), i.e. tan θ=1=tan ϕ, or θ=ϕ=(π/4) θ=ϕ means that the conjugate diameters are equal. (α+β)_(min) =tan^(−1) ((2μ)/(1−μ^2 ))
$$\left({c}\right) \\ $$$${say}\:{the}\:{parameters}\:{of}\:{the}\:{ellipse}\:{are} \\ $$$${a},\:{b}\:{with}\:{a}\geqslant{b}\:{and}\:\mu=\frac{{b}}{{a}}\leqslant\mathrm{1}. \\ $$$${say}\:{P}\left({a}\:\mathrm{cos}\:\theta,\:{b}\:\mathrm{sin}\:\theta\right) \\ $$$$\mathrm{tan}\:\alpha=\frac{{b}\:\mathrm{tan}\:\theta}{{a}}=\mu\:\mathrm{tan}\:\theta \\ $$$$\mathrm{tan}\:\beta=\frac{{b}}{{a}\:\mathrm{tan}\:\theta}=\frac{\mu}{\mathrm{tan}\:\theta} \\ $$$${say}\:{Q}\left({a}\:\mathrm{cos}\:\varphi,\:−{b}\:\mathrm{sin}\:\varphi\right) \\ $$$$\mathrm{tan}\:\beta=\frac{{b}\:\mathrm{tan}\:\varphi}{{a}}=\mu\:\mathrm{tan}\:\varphi \\ $$$$\mathrm{tan}\:\alpha=\frac{{b}}{{a}\:\mathrm{tan}\:\varphi}=\frac{\mu}{\mathrm{tan}\:\varphi} \\ $$$$\Rightarrow\mathrm{tan}\:\theta\:\mathrm{tan}\:\varphi=\mathrm{1} \\ $$$$\Phi=\mathrm{tan}\:\left(\alpha+\beta\right) \\ $$$$\:\:\:=\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta} \\ $$$$\:\:\:=\frac{\mu\:\mathrm{tan}\:\theta+\mu\:\mathrm{tan}\:\varphi}{\mathrm{1}−\mu^{\mathrm{2}} \:\mathrm{tan}\:\theta\:\mathrm{tan}\:\varphi} \\ $$$$\:\:\:=\frac{\mu\left(\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\right)}{\mathrm{1}−\mu^{\mathrm{2}} }\geqslant\frac{\mathrm{2}\mu}{\mathrm{1}−\mu^{\mathrm{2}} } \\ $$$$\Phi_{{min}} =\frac{\mathrm{2}\mu}{\mathrm{1}−\mu^{\mathrm{2}} }\:{when}\:\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\mathrm{tan}\:\theta},\:{i}.{e}. \\ $$$$\mathrm{tan}\:\theta=\mathrm{1}=\mathrm{tan}\:\varphi,\:{or}\:\theta=\varphi=\frac{\pi}{\mathrm{4}} \\ $$$$\theta=\varphi\:{means}\:{that}\:{the}\:{conjugate}\: \\ $$$${diameters}\:{are}\:{equal}. \\ $$$$\left(\alpha+\beta\right)_{{min}} =\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}\mu}{\mathrm{1}−\mu^{\mathrm{2}} } \\ $$
Commented by hardmath last updated on 11/Aug/24
Dear professor, your solutions are perfect
$$ \\ $$Dear professor, your solutions are perfect
Commented by mr W last updated on 11/Aug/24
thanks!
$${thanks}! \\ $$
Commented by peter frank last updated on 14/Aug/24
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by mr W last updated on 11/Aug/24
(c) say the pole is P(u, v). the polar of P with respect to the hyperbola (x^2 /a^2 )−(y^2 /b^2 )=1 is ((ux)/a^2 )−((vy)/b^2 )=1 such that it tangents the circle x^2 +y^2 =r^2 (with r=ae=(√(a^2 +b^2 ))), the distance from (0, 0) to ((ux)/a^2 )−((vy)/b^2 )=1 must be r, i.e. (1/( (√(((u/a^2 ))^2 +(−(v/b^2 ))^2 ))))=r=(√(a^2 +b^2 )) ⇒(u^2 /a^4 )+(v^2 /b^4 )=(1/(a^2 +b^2 )) i.e. the locus of the pole is (x^2 /a^4 )+(y^2 /b^4 )=(1/(a^2 +b^2 ))
$$\left({c}\right) \\ $$$${say}\:{the}\:{pole}\:{is}\:{P}\left({u},\:{v}\right). \\ $$$${the}\:{polar}\:{of}\:{P}\:{with}\:{respect}\:{to}\:{the} \\ $$$${hyperbola}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:{is} \\ $$$$\frac{{ux}}{{a}^{\mathrm{2}} }−\frac{{vy}}{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${such}\:{that}\:{it}\:{tangents}\:{the}\:{circle} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \:\left({with}\:{r}={ae}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right), \\ $$$${the}\:{distance}\:{from}\:\left(\mathrm{0},\:\mathrm{0}\right)\:{to} \\ $$$$\frac{{ux}}{{a}^{\mathrm{2}} }−\frac{{vy}}{{b}^{\mathrm{2}} }=\mathrm{1}\:{must}\:{be}\:{r},\:{i}.{e}. \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\left(\frac{{u}}{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} +\left(−\frac{{v}}{{b}^{\mathrm{2}} }\right)^{\mathrm{2}} }}={r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{u}^{\mathrm{2}} }{{a}^{\mathrm{4}} }+\frac{{v}^{\mathrm{2}} }{{b}^{\mathrm{4}} }=\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${i}.{e}.\:{the}\:{locus}\:{of}\:{the}\:{pole}\:{is} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{4}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{4}} }=\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$
Commented by peter frank last updated on 11/Aug/24
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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