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Question-45639

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Question Number 45639 by peter frank last updated on 14/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18
1)(x^2 /a^2 )+(y^2 /b^2 )=1 (d/dx)((x^2 /a^2 )+(y^2 /b^2 ))=0 ((2x)/a^2 )+((2y)/b^2 )×(dy/dx)=0 (dy/dx)=−(((2x)/a^2 )/((2y)/b^2 ))=−((b^2 x)/(a^2 y)) so slope of tangent at point(acosθ,bsinθ) m=−((b^2 ×acosθ)/(a^2 ×bsinθ))=−((bcosθ)/(asinθ)) so eqn of tangeng is (y−bsinθ)=−((bcosθ)/(asinθ))(x−acosθ) yasinθ−absin^2 θ+xbcosθ−abcos^2 θ=0 yasinθ+xbcosθ=ab ((ysinθ)/b)+((xcosθ)/a)=1 2)((xcosθ)/a)+((ysinθ)/b)=1 (x/(a/(cosθ)))+(y/(b/(sinθ)))=1 compare this eqn with straightt line in intercepted form (x/A)+(y/B)=1 A=(a/(cosθ))=intercept in x axis B=(b/(sinθ)) intercept on y axis so area of triangle (1/2)×(a/(cosθ))×(b/(sinθ)) =((ab)/(sin2θ))
$$\left.\mathrm{1}\right)\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{d}}{{dx}}\left(\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$ \\ $$$$\frac{\mathrm{2}{x}}{{a}^{\mathrm{2}} }+\frac{\mathrm{2}{y}}{{b}^{\mathrm{2}} }×\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=−\frac{\frac{\mathrm{2}{x}}{{a}^{\mathrm{2}} }}{\frac{\mathrm{2}{y}}{{b}^{\mathrm{2}} }}=−\frac{{b}^{\mathrm{2}} {x}}{{a}^{\mathrm{2}} {y}} \\ $$$${so}\:{slope}\:{of}\:{tangent}\:{at}\:{point}\left({acos}\theta,{bsin}\theta\right) \\ $$$${m}=−\frac{{b}^{\mathrm{2}} ×{acos}\theta}{{a}^{\mathrm{2}} ×{bsin}\theta}=−\frac{{bcos}\theta}{{asin}\theta} \\ $$$${so}\:{eqn}\:{of}\:{tangeng}\:{is}\: \\ $$$$\left({y}−{bsin}\theta\right)=−\frac{{bcos}\theta}{{asin}\theta}\left({x}−{acos}\theta\right) \\ $$$${yasin}\theta−{absin}^{\mathrm{2}} \theta+{xbcos}\theta−{abcos}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$${yasin}\theta+{xbcos}\theta={ab} \\ $$$$\frac{{ysin}\theta}{{b}}+\frac{{xcos}\theta}{{a}}=\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\frac{{xcos}\theta}{{a}}+\frac{{ysin}\theta}{{b}}=\mathrm{1} \\ $$$$\frac{{x}}{\frac{{a}}{{cos}\theta}}+\frac{{y}}{\frac{{b}}{{sin}\theta}}=\mathrm{1}\:\: \\ $$$${compare}\:{this}\:{eqn}\:{with}\:{straightt}\:{line}\:{in}\:{intercepted} \\ $$$${form}\:\:\: \\ $$$$\frac{{x}}{{A}}+\frac{{y}}{{B}}=\mathrm{1}\:\:\:{A}=\frac{{a}}{{cos}\theta}={intercept}\:{in}\:{x}\:{axis} \\ $$$${B}=\frac{{b}}{{sin}\theta}\:{intercept}\:{on}\:{y}\:{axis} \\ $$$${so}\:{area}\:{of}\:{triangle} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{{cos}\theta}×\frac{{b}}{{sin}\theta} \\ $$$$=\frac{{ab}}{{sin}\mathrm{2}\theta} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18
3)A(a,0) B(0,b) P(acosθ,bsinθ) area of triangle formula (1/2)∣{x_1 (y_2 −y_3 )+x_2 (y_3 −y_1 )+x_3 (y_1 −y_2 )}∣ =(1/2)∣{a(b−bsinθ)+0(bsinθ−0)+acosθ(0−b)}∣ =(1/2)∣{ab−absinθ−abcosθ}∣
$$\left.\mathrm{3}\right){A}\left({a},\mathrm{0}\right)\:\:{B}\left(\mathrm{0},{b}\right)\:\:{P}\left({acos}\theta,{bsin}\theta\right) \\ $$$${area}\:{of}\:{triangle}\:{formula} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mid\left\{{x}_{\mathrm{1}} \left({y}_{\mathrm{2}} −{y}_{\mathrm{3}} \right)+{x}_{\mathrm{2}} \left({y}_{\mathrm{3}} −{y}_{\mathrm{1}} \right)+{x}_{\mathrm{3}} \left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)\right\}\mid \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mid\left\{{a}\left({b}−{bsin}\theta\right)+\mathrm{0}\left({bsin}\theta−\mathrm{0}\right)+{acos}\theta\left(\mathrm{0}−{b}\right)\right\}\mid \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mid\left\{{ab}−{absin}\theta−{abcos}\theta\right\}\mid \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18
d)area of APB=(1/2)ab(cosθ+sinθ−1) already proved S=((ab)/2)(cosθ+sinθ−1) (dS/dθ)=((ab)/2)(−sinθ+cosθ) for max/min (dS/dθ)=0=((ab)/2)(−sinθ+cosθ) sinθ=cosθ so tanθ=1=tan(π/4) θ=(π/4) (d^2 S/dθ^2 )=((ab)/2)(−sinθ−cosθ)=−((ab)/2)(sinθ+cosθ) in the interval 0<θ<(π/2) sinθ+cosθ=+ve so(d^2 θ/dθ^2 )=−ve now point A=(a,0) B=(0,b) slope of AB m_1 =((b−0)/(0−a))=−(b/a) tangent at p is (already proved) ((xcosθ)/a)+((ysinθ)/b)=1 at θ=(π/4) (x/(a(√2)))+(y/(b(√2)))=1 (y/(b(√2)))=1−(x/(a(√2))) y=b(√2) −(b/a)x slope is m_2 =−(b/a) so m_1 =m_2 =((−b)/a) that means AB is parallel to tangent at p atθ=(π/4) pls check...
$$\left.{d}\right){area}\:{of}\:{APB}=\frac{\mathrm{1}}{\mathrm{2}}{ab}\left({cos}\theta+{sin}\theta−\mathrm{1}\right) \\ $$$${already}\:{proved} \\ $$$${S}=\frac{{ab}}{\mathrm{2}}\left({cos}\theta+{sin}\theta−\mathrm{1}\right) \\ $$$$\frac{{dS}}{{d}\theta}=\frac{{ab}}{\mathrm{2}}\left(−{sin}\theta+{cos}\theta\right) \\ $$$$ \\ $$$${for}\:{max}/{min}\:\frac{{dS}}{{d}\theta}=\mathrm{0}=\frac{{ab}}{\mathrm{2}}\left(−{sin}\theta+{cos}\theta\right) \\ $$$${sin}\theta={cos}\theta\:\:\:{so}\:\:{tan}\theta=\mathrm{1}={tan}\frac{\pi}{\mathrm{4}}\:\:\:\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$$\frac{{d}^{\mathrm{2}} {S}}{{d}\theta^{\mathrm{2}} }=\frac{{ab}}{\mathrm{2}}\left(−{sin}\theta−{cos}\theta\right)=−\frac{{ab}}{\mathrm{2}}\left({sin}\theta+{cos}\theta\right) \\ $$$${in}\:{the}\:{interval}\:\:\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:\: \\ $$$${sin}\theta+{cos}\theta=+{ve}\:\:\:{so}\frac{{d}^{\mathrm{2}} \theta}{{d}\theta^{\mathrm{2}} }=−{ve}\: \\ $$$${now}\:{point}\:{A}=\left({a},\mathrm{0}\right)\:\:\:{B}=\left(\mathrm{0},{b}\right) \\ $$$${slope}\:{of}\:\:{AB}\:{m}_{\mathrm{1}} =\frac{{b}−\mathrm{0}}{\mathrm{0}−{a}}=−\frac{{b}}{{a}} \\ $$$${tangent}\:{at}\:{p}\:{is}\:\left({already}\:{proved}\right) \\ $$$$\frac{{xcos}\theta}{{a}}+\frac{{ysin}\theta}{{b}}=\mathrm{1}\:\:{at}\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$$\frac{{x}}{{a}\sqrt{\mathrm{2}}}+\frac{{y}}{{b}\sqrt{\mathrm{2}}}=\mathrm{1}\:\:\:\:\frac{{y}}{{b}\sqrt{\mathrm{2}}}=\mathrm{1}−\frac{{x}}{{a}\sqrt{\mathrm{2}}} \\ $$$${y}={b}\sqrt{\mathrm{2}}\:−\frac{{b}}{{a}}{x}\:\:\: \\ $$$${slope}\:{is}\:\:{m}_{\mathrm{2}} =−\frac{{b}}{{a}} \\ $$$${so}\:{m}_{\mathrm{1}} ={m}_{\mathrm{2}} =\frac{−{b}}{{a}}\:\:\:{that}\:\:{means}\:{AB}\:{is}\:{parallel} \\ $$$${to}\:{tangent}\:{at}\:{p}\:{at}\theta=\frac{\pi}{\mathrm{4}} \\ $$$${pls}\:{check}… \\ $$
Commented by peter frank last updated on 19/Oct/18
the answer given is (1/2)abcosec𝛉
$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{ans}}\mathrm{w}\boldsymbol{\mathrm{er}}\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{is}}\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{abcosec}\theta} \\ $$

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