Question-87952

Question Number 87952 by ajfour last updated on 07/Apr/20

Commented by ajfour last updated on 07/Apr/20

$${what}\:{is}\:{c}\:?\:\:{Sir}. \\ $$

Commented by ajfour last updated on 07/Apr/20

$${Find}\:{equation}\:{of}\:{line}\:{ABCD}\:{in} \\ $$$${terms}\:{of}\:{ellipse}\:{parameters} \\ $$$${a}\:{and}\:{b}. \\ $$

Commented by Prithwish Sen 1 last updated on 07/Apr/20

I got y−c = (b/a).(√(((b^2 −c^2 )/(9c^2 −b^2 )) )).x please check

$$\mathrm{I}\:\mathrm{got} \\ $$$$\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{c}}\:=\:\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}}.\sqrt{\frac{\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\boldsymbol{\mathrm{c}}^{\mathrm{2}} }{\mathrm{9}\boldsymbol{\mathrm{c}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} }\:}.\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$

Commented by john santu last updated on 07/Apr/20

$$\mathrm{c}\:=\:\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} } \\ $$

Commented by Prithwish Sen 1 last updated on 07/Apr/20

D(acosθ,bsinθ) A(acosα, −bsinα) Considering C 2sinθ − sinα =0...i and considering B cosθ +2cosα = 0...ii the equation is y−bsinθ = ((b(sinθ+sinα))/(a(cosθ−cosα))) (x−axcosθ)...iii solving the three equations (x/(2a)) − (y/b) = (1/( (√5))) sir please check.

$$\mathrm{D}\left(\mathrm{acos}\theta,\mathrm{bsin}\theta\right)\:\:\:\mathrm{A}\left(\mathrm{acos}\alpha,\:−\mathrm{bsin}\alpha\right) \\ $$$$\mathrm{Considering}\:\mathrm{C} \\ $$$$\mathrm{2sin}\theta\:−\:\mathrm{sin}\alpha\:=\mathrm{0}…\mathrm{i} \\ $$$$\mathrm{and}\:\mathrm{considering}\:\mathrm{B}\:\:\:\mathrm{cos}\theta\:+\mathrm{2cos}\alpha\:=\:\mathrm{0}…\mathrm{ii} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{is} \\ $$$$\mathrm{y}−\mathrm{bsin}\theta\:=\:\frac{\mathrm{b}\left(\mathrm{sin}\theta+\mathrm{sin}\alpha\right)}{\mathrm{a}\left(\mathrm{cos}\theta−\mathrm{cos}\alpha\right)}\:\left(\mathrm{x}−\mathrm{axcos}\theta\right)…\mathrm{iii} \\ $$$$\mathrm{solving}\:\mathrm{the}\:\mathrm{three}\:\mathrm{equations}\: \\ $$$$\:\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}\boldsymbol{\mathrm{a}}}\:−\:\frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{b}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$$$ \\ $$$$ \\ $$

Answered by ajfour last updated on 07/Apr/20

Let eq. of line_(−) be (x/p)−(y/q)=1 , slope m=(q/p) l=(√(p^2 +q^2 )) D≡(2p,q) ; A≡(−p,−2q) ((4p^2 )/a^2 )+(q^2 /b^2 )=1 & (p^2 /a^2 )+((4q^2 )/b^2 )=1 ⇒ ((15p^2 )/a^2 )=3 ⇒ p=(a/( (√5))) & ((15q^2 )/b^2 )=3 ⇒ q=(b/( (√5))) ⇒ eq. of line is (x/a)−(y/b)=(1/( (√5))) .

$${Let}\:\underset{−} {{eq}.\:{of}\:{line}}\:{be} \\ $$$$\frac{{x}}{{p}}−\frac{{y}}{{q}}=\mathrm{1}\:\:\:,\:{slope}\:{m}=\frac{{q}}{{p}} \\ $$$${l}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} } \\ $$$${D}\equiv\left(\mathrm{2}{p},{q}\right)\:\:\:;\:\:{A}\equiv\left(−{p},−\mathrm{2}{q}\right) \\ $$$$\frac{\mathrm{4}{p}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{q}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:\&\:\:\frac{{p}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{4}{q}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{\mathrm{15}{p}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{3}\:\:\Rightarrow\:\:{p}=\frac{{a}}{\:\sqrt{\mathrm{5}}} \\ $$$$\&\:\:\:\frac{\mathrm{15}{q}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{3}\:\:\Rightarrow\:\:{q}=\frac{{b}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow\:\:{eq}.\:{of}\:{line}\:{is} \\ $$$$\:\:\:\:\:\frac{{x}}{{a}}−\frac{{y}}{{b}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:. \\ $$

Commented by ajfour last updated on 07/Apr/20

$${Sen}\:{Sir},\:{please}\:{check}\:{my}\:{sol}^{{n}} . \\ $$

Commented by Prithwish Sen 1 last updated on 07/Apr/20

sir yours is right but please check mine. I cannot find any wrong in my calculation. please help

$$\mathrm{sir}\:\mathrm{yours}\:\mathrm{is}\:\mathrm{right}\:\mathrm{but}\:\mathrm{please}\:\mathrm{check}\:\mathrm{mine}.\:\mathrm{I}\:\mathrm{cannot} \\ $$$$\mathrm{find}\:\mathrm{any}\:\mathrm{wrong}\:\mathrm{in}\:\mathrm{my}\:\mathrm{calculation}.\:\mathrm{please}\:\mathrm{help} \\ $$

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