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Question Number 194064 by Abdullahrussell last updated on 27/Jun/23

Answered by som(math1967) last updated on 27/Jun/23

i)×asecθ −ii)×((cosθ)/a) ((ay)/b)tanθ +((by)/a)cotθ=asecθ−((cosθ)/a)+((b^2 cosθ)/a) y(((a^2 tan^2 θ+b^2 )/(abtanθ)))=((a^2 (sec^2 θ−1)+b^2 )/(asecθ)) y(((a^2 tan^2 θ+b^2 )/(abtanθ)))=(((a^2 tan^2 θ+b^2 ))/(asecθ)) ⇒y=bsinθ ∴ (y/b)=sinθ putting y=bsinθ in (i) ((xcos θ)/a) +sin^2 θ=1 ⇒x=acosθ ∴(x/a)=cosθ ∴(x^2 /a^2 )+(y^2 /b^2 )=cos^2 θ+sin^2 θ=1

$$\left.{i}\left.\right)×{asec}\theta\:−{ii}\right)×\frac{{cos}\theta}{{a}} \\ $$$$\:\frac{{ay}}{{b}}{tan}\theta\:+\frac{{by}}{{a}}{cot}\theta={asec}\theta−\frac{{cos}\theta}{{a}}+\frac{{b}^{\mathrm{2}} {cos}\theta}{{a}} \\ $$$${y}\left(\frac{{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta+{b}^{\mathrm{2}} }{{abtan}\theta}\right)=\frac{{a}^{\mathrm{2}} \left({sec}^{\mathrm{2}} \theta−\mathrm{1}\right)+{b}^{\mathrm{2}} }{{asec}\theta} \\ $$$$\:{y}\left(\frac{{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta+{b}^{\mathrm{2}} }{{abtan}\theta}\right)=\frac{\left({a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta+{b}^{\mathrm{2}} \right)}{{asec}\theta} \\ $$$$\:\Rightarrow{y}={bsin}\theta \\ $$$$\therefore\:\frac{{y}}{{b}}={sin}\theta \\ $$$$\:{putting}\:{y}={bsin}\theta\:\:{in}\:\left({i}\right) \\ $$$$\:\frac{{x}\mathrm{cos}\:\theta}{{a}}\:+{sin}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\Rightarrow{x}={acos}\theta \\ $$$$\therefore\frac{{x}}{{a}}={cos}\theta \\ $$$$\therefore\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }={cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta=\mathrm{1} \\ $$

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