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xcos-a-ysin-b-1-xsin-ycos-a-2-sin-2-b-2-cos-2-Eliminate-

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Question Number 207594 by MATHEMATICSAM last updated on 19/May/24
((xcosθ)/a) + ((ysinθ)/b) = 1 xsinθ − ycosθ = (√(a^2 sin^2 θ + b^2 cos^2 θ)) Eliminate θ.
$$\frac{{x}\mathrm{cos}\theta}{{a}}\:+\:\frac{{y}\mathrm{sin}\theta}{{b}}\:=\:\mathrm{1} \\ $$$${x}\mathrm{sin}\theta\:−\:{y}\mathrm{cos}\theta\:=\:\sqrt{{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta\:+\:{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\mathrm{Eliminate}\:\theta. \\ $$
Answered by Frix last updated on 20/May/24
t=tan θ and transforming gives A. (x/a)+((ty)/b)=(√(t^2 +1)) [⇒ (x/a)+((ty)/b)≥0] B. tx−y=(√(a^2 t^2 +b^2 )) [⇒ tx≥y] Squaring [might introduce false solutions!] and transforming gives t^2 +((2bxy)/(a(y^2 −b^2 )))t+((b^2 (x^2 −a^2 ))/(a^2 (y^2 −b^2 )))=0 [⇒ y≠±b] t^2 −((2xy)/(x^2 −a^2 ))t+((y^2 −b^2 )/(x^2 −a^2 ))=0 [⇒ x≠±a] Subtracting and solving gives t=−((bx^2 −ay^2 −ab(a−b))/(2axy)) Now insert this in A or B I′m not sure if this is a valid solution. The given equations define two lines in R^2 with a, b, θ as constants or rather strange objects in R^3 with a, b as constants...
$${t}=\mathrm{tan}\:\theta\:\mathrm{and}\:\mathrm{transforming}\:\mathrm{gives} \\ $$$${A}.\:\frac{{x}}{{a}}+\frac{{ty}}{{b}}=\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\:\:\:\:\:\:\:\:\left[\Rightarrow\:\frac{{x}}{{a}}+\frac{{ty}}{{b}}\geqslant\mathrm{0}\right] \\ $$$${B}.\:{tx}−{y}=\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:\:\:\:\left[\Rightarrow\:{tx}\geqslant{y}\right] \\ $$$$\mathrm{Squaring}\:\left[\mathrm{might}\:\mathrm{introduce}\:\mathrm{false}\:\mathrm{solutions}!\right] \\ $$$$\mathrm{and}\:\mathrm{transforming}\:\mathrm{gives} \\ $$$${t}^{\mathrm{2}} +\frac{\mathrm{2}{bxy}}{{a}\left({y}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{t}+\frac{{b}^{\mathrm{2}} \left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} \left({y}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}=\mathrm{0}\:\:\:\:\:\left[\Rightarrow\:{y}\neq\pm{b}\right] \\ $$$${t}^{\mathrm{2}} −\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }{t}+\frac{{y}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\Rightarrow\:{x}\neq\pm{a}\right] \\ $$$$\mathrm{Subtracting}\:\mathrm{and}\:\mathrm{solving}\:\mathrm{gives} \\ $$$${t}=−\frac{{bx}^{\mathrm{2}} −{ay}^{\mathrm{2}} −{ab}\left({a}−{b}\right)}{\mathrm{2}{axy}} \\ $$$$\mathrm{Now}\:\mathrm{insert}\:\mathrm{this}\:\mathrm{in}\:{A}\:\mathrm{or}\:{B} \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{valid}\:\mathrm{solution}.\:\mathrm{The} \\ $$$$\mathrm{given}\:\mathrm{equations}\:\mathrm{define}\:\mathrm{two}\:\mathrm{lines}\:\mathrm{in}\:\mathbb{R}^{\mathrm{2}} \:\mathrm{with} \\ $$$${a},\:{b},\:\theta\:\mathrm{as}\:\mathrm{constants}\:\mathrm{or}\:\mathrm{rather}\:\mathrm{strange}\:\mathrm{objects} \\ $$$$\mathrm{in}\:\mathbb{R}^{\mathrm{3}} \:\mathrm{with}\:{a},\:{b}\:\mathrm{as}\:\mathrm{constants}… \\ $$
Answered by mr W last updated on 20/May/24
(x/a) cos θ+(y/b) sin θ=1 cos α cos θ+sin α sin θ=(1/( (√((x^2 /a^2 )+(y^2 /b^2 ))))) cos (θ−α)=(1/( (√((x^2 /a^2 )+(y^2 /b^2 ))))) θ=α+cos^(−1) (1/( (√((x^2 /a^2 )+(y^2 /b^2 ))))) θ=cos^(−1) (x/(a(√((x^2 /a^2 )+(y^2 /b^2 )))))+cos^(−1) (1/( (√((x^2 /a^2 )+(y^2 /b^2 ))))) ⇒sin θ=(y/( b((x^2 /a^2 )+(y^2 /b^2 ))))+((x(√(1−(x^2 /a^2 )−(y^2 /b^2 ))))/( a((x^2 /a^2 )+(y^2 /b^2 )))) ⇒cos θ=(x/( a((x^2 /a^2 )+(y^2 /b^2 ))))−((y(√(1−(x^2 /a^2 )−(y^2 /b^2 ))))/( b((x^2 /a^2 )+(y^2 /b^2 )))) insert this into equation (ii)
$$\frac{{x}}{{a}}\:\mathrm{cos}\:\theta+\frac{{y}}{{b}}\:\mathrm{sin}\:\theta=\mathrm{1} \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{cos}\:\theta+\mathrm{sin}\:\alpha\:\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}} \\ $$$$\mathrm{cos}\:\left(\theta−\alpha\right)=\frac{\mathrm{1}}{\:\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}} \\ $$$$\theta=\alpha+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}} \\ $$$$\theta=\mathrm{cos}^{−\mathrm{1}} \frac{{x}}{{a}\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}}+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{{y}}{\:{b}\left(\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)}+\frac{{x}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}}{\:{a}\left(\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{{x}}{\:{a}\left(\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)}−\frac{{y}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}}{\:{b}\left(\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)} \\ $$$${insert}\:{this}\:{into}\:{equation}\:\left({ii}\right) \\ $$

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