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Question-90326




Question Number 90326 by ajfour last updated on 22/Apr/20
Commented by ajfour last updated on 22/Apr/20
Find a/b. Ellipse  b^2 x^2 +a^2 y^2 =a^2 b^2 .
$${Find}\:{a}/{b}.\:{Ellipse}\:\:{b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} . \\ $$
Answered by ajfour last updated on 22/Apr/20
y=x−b  b^2 x^2 +a^2 (x−b)^2 =a^2 b^2   eq. above has two roots  x=x_1 , x=0  x_1 =((2a^2 b)/(a^2 +b^2 ))  ,  y_1 =(((a^2 −b^2 )b)/(a^2 +b^2 ))    (dy/dx)∣_(x=x_1 ) = −(b^2 /a^2 )((x_1 /y_1 ))=−1  ⇒   (b^2 /a^2 )(((2a^2 )/(a^2 −b^2 )))=1  ⇒   (a^2 /b^2 ) =3    ⇒   (a/b)=(√3) .
$${y}={x}−{b} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({x}−{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${eq}.\:{above}\:{has}\:{two}\:{roots} \\ $$$${x}={x}_{\mathrm{1}} ,\:{x}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{2}{a}^{\mathrm{2}} {b}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:,\:\:{y}_{\mathrm{1}} =\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){b}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}\mid_{{x}={x}_{\mathrm{1}} } =\:−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\left(\frac{{x}_{\mathrm{1}} }{{y}_{\mathrm{1}} }\right)=−\mathrm{1} \\ $$$$\Rightarrow\:\:\:\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\left(\frac{\mathrm{2}{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\mathrm{3}\:\:\:\:\Rightarrow\:\:\:\frac{{a}}{{b}}=\sqrt{\mathrm{3}}\:. \\ $$
Commented by ajfour last updated on 22/Apr/20
tough one for me, i′ll try  tomorrow.
$${tough}\:{one}\:{for}\:{me},\:{i}'{ll}\:{try} \\ $$$${tomorrow}. \\ $$

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