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Question Number 50638 by ajfour last updated on 18/Dec/18

Commented by ajfour last updated on 18/Dec/18

Determine (a/b) for ellipse if α = 45° .

$${Determine}\:\frac{{a}}{{b}}\:{for}\:{ellipse}\:{if}\:\alpha\:=\:\mathrm{45}°\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Dec/18

eqn normal y=xtanα eqn ellipse (((x−0)^2 )/a^2 )+(((y−b)^2 )/b^2 )=1 acosθ,b+bsinθ lies on normal so b+bsinθ=acosθtanα b^2 x^2 +a^2 (y^2 −2yb+b^2 )=a^2 b^2 b^2 x^2 +a^2 y^2 −2a^2 by+a^2 b^2 =a^2 b^2 eqn tangent b^2 x(acosθ)+a^2 y(b+bsinθ)−a^2 b(y+b+bsinθ)=0 x(ab^2 cosθ)+y(a^2 b+a^2 bsinθ−a^2 b)=a^2 b(b+bsinθ) slope=m=((−ab^2 cosθ)/(a^2 bsinθ))=−(b/a)cotθ←slope of tangent slope of normal=(a/(bcotθ)) tanα=(a/(bcotθ)) l=(√((acosθ)^2 +(b+bsinθ)^2 )) =(√(a^2 cos^2 θ+b^2 +2b^2 sinθ+b^2 sin^2 θ)) lcosα=acosθ lsinα=b+binθ tanα=((b+bsinθ)/(acosθ))=(a/(bcotθ)) ((b+bsinθ)/a)=((asinθ)/b) a^2 sinθ=b^2 +b^2 sinθ sinθ=(b^2 /(a^2 −b^2 )) cotθ=((base)/(perpendicular))=((√((a^2 −b^2 )^2 −b^4 ))/b^2 ) cotθ=((√(a^4 −2a^2 b^2 ))/b^2 ) now tanα=(a/(bcotθ))=(a/b)×(b^2 /(√(a^4 −2a^2 b^2 ))) tanα=(a/b)×(b^2 /a)×(1/(√(a^2 −2b^2 )))=(b/(b(√(((a/b))^2 −2)))) tanα=(1/(√(((a/b))^2 −2))) given α=(π/4) (√(((a/b))^2 −2)) =1 ((a/b))^2 =3 so (a/b)=(√3) pls check...

$${eqn}\:{normal}\:{y}={xtan}\alpha \\ $$$${eqn}\:{ellipse} \\ $$$$\frac{\left({x}−\mathrm{0}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({y}−{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${acos}\theta,{b}+{bsin}\theta\:{lies}\:{on}\:{normal} \\ $$$${so} \\ $$$${b}+{bsin}\theta={acos}\theta{tan}\alpha \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({y}^{\mathrm{2}} −\mathrm{2}{yb}+{b}^{\mathrm{2}} \right)={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {by}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${eqn}\:{tangent} \\ $$$${b}^{\mathrm{2}} {x}\left({acos}\theta\right)+{a}^{\mathrm{2}} {y}\left({b}+{bsin}\theta\right)−{a}^{\mathrm{2}} {b}\left({y}+{b}+{bsin}\theta\right)=\mathrm{0} \\ $$$${x}\left({ab}^{\mathrm{2}} {cos}\theta\right)+{y}\left({a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {bsin}\theta−{a}^{\mathrm{2}} {b}\right)={a}^{\mathrm{2}} {b}\left({b}+{bsin}\theta\right) \\ $$$${slope}={m}=\frac{−{ab}^{\mathrm{2}} {cos}\theta}{{a}^{\mathrm{2}} {bsin}\theta}=−\frac{{b}}{{a}}{cot}\theta\leftarrow{slope}\:{of}\:{tangent} \\ $$$${slope}\:{of}\:{normal}=\frac{{a}}{{bcot}\theta} \\ $$$${tan}\alpha=\frac{{a}}{{bcot}\theta} \\ $$$${l}=\sqrt{\left({acos}\theta\right)^{\mathrm{2}} +\left({b}+{bsin}\theta\right)^{\mathrm{2}} }\:=\sqrt{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta+{b}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} {sin}\theta+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}\: \\ $$$${lcos}\alpha={acos}\theta \\ $$$${lsin}\alpha={b}+{bin}\theta \\ $$$$ \\ $$$${tan}\alpha=\frac{{b}+{bsin}\theta}{{acos}\theta}=\frac{{a}}{{bcot}\theta} \\ $$$$\frac{{b}+{bsin}\theta}{{a}}=\frac{{asin}\theta}{{b}} \\ $$$${a}^{\mathrm{2}} {sin}\theta={b}^{\mathrm{2}} +{b}^{\mathrm{2}} {sin}\theta \\ $$$${sin}\theta=\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${cot}\theta=\frac{{base}}{{perpendicular}}=\frac{\sqrt{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} −{b}^{\mathrm{4}} }}{{b}^{\mathrm{2}} } \\ $$$${cot}\theta=\frac{\sqrt{{a}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} } \\ $$$${now}\:{tan}\alpha=\frac{{a}}{{bcot}\theta}=\frac{{a}}{{b}}×\frac{{b}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }} \\ $$$${tan}\alpha=\frac{{a}}{{b}}×\frac{{b}^{\mathrm{2}} }{{a}}×\frac{\mathrm{1}}{\sqrt{{a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} }}=\frac{{b}}{{b}\sqrt{\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{2}}} \\ $$$${tan}\alpha=\frac{\mathrm{1}}{\sqrt{\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{2}}} \\ $$$${given}\:\alpha=\frac{\pi}{\mathrm{4}} \\ $$$$\sqrt{\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{2}}\:=\mathrm{1} \\ $$$$\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} =\mathrm{3}\:\:\:{so}\:\frac{{a}}{{b}}=\sqrt{\mathrm{3}}\:\:{pls}\:{check}... \\ $$$$ \\ $$

Commented by MJS last updated on 18/Dec/18

I get b=a((√3)/3) ⇒ (a/b)=(√3)

$$\mathrm{I}\:\mathrm{get}\:{b}={a}\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:\Rightarrow\:\frac{{a}}{{b}}=\sqrt{\mathrm{3}} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 18/Dec/18

thank you sir i got the same result...

$${thank}\:{you}\:{sir}\:{i}\:{got}\:{the}\:{same}\:{result}... \\ $$

Answered by mr W last updated on 18/Dec/18

P(h,k) ⇒k=h (h^2 /a^2 )+(((h−b)^2 )/b^2 )=1 (h/a^2 )+((h−b)/b^2 )(−1)=0 (h^2 /a^2 )=(((h−b)h)/b^2 ) (((h−b)h)/b^2 )+(((h−b)^2 )/b^2 )=1 ⇒h=((3b)/2) ((9b^2 )/(4a^2 ))+(b^2 /(4b^2 ))=1 (b^2 /a^2 )=(1/3) ⇒(b/a)=(1/(√3))≈0.577

$${P}\left({h},{k}\right) \\ $$$$\Rightarrow{k}={h} \\ $$$$\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({h}−{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{h}}{{a}^{\mathrm{2}} }+\frac{{h}−{b}}{{b}^{\mathrm{2}} }\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\frac{\left({h}−{b}\right){h}}{{b}^{\mathrm{2}} } \\ $$$$\frac{\left({h}−{b}\right){h}}{{b}^{\mathrm{2}} }+\frac{\left({h}−{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{h}=\frac{\mathrm{3}{b}}{\mathrm{2}} \\ $$$$\frac{\mathrm{9}{b}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }+\frac{{b}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{b}}{{a}}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\approx\mathrm{0}.\mathrm{577} \\ $$

Commented by mr W last updated on 18/Dec/18

(x^2 /a^2 )+(((y−b)^2 )/b^2 )=1 (x/a^2 )+(((y−b))/b^2 )y′=0 at (h,k): y′=−1 ⇒(h/a^2 )+(((k−b))/b^2 )(−1)=0

$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({y}−{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}}{{a}^{\mathrm{2}} }+\frac{\left({y}−{b}\right)}{{b}^{\mathrm{2}} }{y}'=\mathrm{0} \\ $$$${at}\:\left({h},{k}\right):\:{y}'=−\mathrm{1} \\ $$$$\Rightarrow\frac{{h}}{{a}^{\mathrm{2}} }+\frac{\left({k}−{b}\right)}{{b}^{\mathrm{2}} }\left(−\mathrm{1}\right)=\mathrm{0} \\ $$

Commented by ajfour last updated on 18/Dec/18

Sir please explain 4th line (h/a^2 )+(((h−b))/b^2 )(−1) = 0

$${Sir}\:{please}\:{explain}\:\:\mathrm{4}{th}\:{line} \\ $$$$\:\frac{{h}}{{a}^{\mathrm{2}} }+\frac{\left({h}−{b}\right)}{{b}^{\mathrm{2}} }\left(−\mathrm{1}\right)\:=\:\mathrm{0} \\ $$

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