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Question Number 213281 by ajfour last updated on 02/Nov/24
Commented by ajfour last updated on 02/Nov/24
Find r in terms of a, b.
$$\:\:\:\:{Find}\:{r}\:{in}\:{terms}\:{of}\:{a},\:{b}. \\ $$
Answered by mr W last updated on 02/Nov/24
say touching point from circle and parabola at (±p, p^2 ). tan θ=2p r sin θ=p ⇒r=((√(1+4p^2 ))/( 2)) y_c =p^2 +r cos θ=p^2 +(1/2) eqn. of tangent line: y=−((b^2 (x−a))/(a+b))=((ab^2 )/(a+b))−((b^2 x)/(a+b)) x^2 +(((ab^2 )/(a+b))−((b^2 x)/(a+b))−p^2 −(1/2))^2 =((1+4p^2 )/4) x^2 +(((ab^2 )/(a+b))−p^2 −(1/2)−((b^2 x)/(a+b)))^2 =((1+4p^2 )/4) [1+(b^4 /((a+b)^2 ))]x^2 −((2b^2 )/((a+b)))(((ab^2 )/(a+b))−p^2 −(1/2))x+(((ab^2 )/(a+b))−p^2 −(1/2))^2 −((1+4p^2 )/4)=0 Δ=(b^4 /((a+b)^2 ))(((ab^2 )/(a+b))−p^2 −(1/2))^2 −[1+(b^4 /((a+b)^2 ))][(((ab^2 )/(a+b))−p^2 −(1/2))^2 −((1+4p^2 )/4)]=0 ((1/4)+p^2 )[1+(b^4 /((a+b)^2 ))]−(((ab^2 )/(a+b))−(1/2)−p^2 )^2 =0 p^4 −(b^2 /(a+b))(2a+(b^2 /(a+b)))p^2 +(a^2 +(b^2 /4))((a^2 b^4 )/((a+b)^2 ))−((ab^2 )/(a+b))=0 ⇒p^2 =(1/2){(b^2 /(a+b))(2a+(b^2 /(a+b)))−(√([(2ab+(b^3 /(a+b)))^2 −4a^4 −a^2 b^2 ](b^4 /((a+b)^2 ))+((4ab^2 )/(a+b))))}
$${say}\:{touching}\:{point}\:{from}\:{circle}\:{and} \\ $$$${parabola}\:{at}\:\left(\pm{p},\:{p}^{\mathrm{2}} \right). \\ $$$$\mathrm{tan}\:\theta=\mathrm{2}{p} \\ $$$${r}\:\mathrm{sin}\:\theta={p}\: \\ $$$$\Rightarrow{r}=\frac{\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}{\:\mathrm{2}} \\ $$$${y}_{{c}} ={p}^{\mathrm{2}} +{r}\:\mathrm{cos}\:\theta={p}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${eqn}.\:{of}\:{tangent}\:{line}: \\ $$$${y}=−\frac{{b}^{\mathrm{2}} \left({x}−{a}\right)}{{a}+{b}}=\frac{{ab}^{\mathrm{2}} }{{a}+{b}}−\frac{{b}^{\mathrm{2}} {x}}{{a}+{b}} \\ $$$${x}^{\mathrm{2}} +\left(\frac{{ab}^{\mathrm{2}} }{{a}+{b}}−\frac{{b}^{\mathrm{2}} {x}}{{a}+{b}}−{p}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} +\left(\frac{{ab}^{\mathrm{2}} }{{a}+{b}}−{p}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}−\frac{{b}^{\mathrm{2}} {x}}{{a}+{b}}\right)^{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\left[\mathrm{1}+\frac{{b}^{\mathrm{4}} }{\left({a}+{b}\right)^{\mathrm{2}} }\right]{x}^{\mathrm{2}} −\frac{\mathrm{2}{b}^{\mathrm{2}} }{\left({a}+{b}\right)}\left(\frac{{ab}^{\mathrm{2}} }{{a}+{b}}−{p}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right){x}+\left(\frac{{ab}^{\mathrm{2}} }{{a}+{b}}−{p}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\Delta=\frac{{b}^{\mathrm{4}} }{\left({a}+{b}\right)^{\mathrm{2}} }\left(\frac{{ab}^{\mathrm{2}} }{{a}+{b}}−{p}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left[\mathrm{1}+\frac{{b}^{\mathrm{4}} }{\left({a}+{b}\right)^{\mathrm{2}} }\right]\left[\left(\frac{{ab}^{\mathrm{2}} }{{a}+{b}}−{p}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }{\mathrm{4}}\right]=\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{4}}+{p}^{\mathrm{2}} \right)\left[\mathrm{1}+\frac{{b}^{\mathrm{4}} }{\left({a}+{b}\right)^{\mathrm{2}} }\right]−\left(\frac{{ab}^{\mathrm{2}} }{{a}+{b}}−\frac{\mathrm{1}}{\mathrm{2}}−{p}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${p}^{\mathrm{4}} −\frac{{b}^{\mathrm{2}} }{{a}+{b}}\left(\mathrm{2}{a}+\frac{{b}^{\mathrm{2}} }{{a}+{b}}\right){p}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{\mathrm{4}}\right)\frac{{a}^{\mathrm{2}} {b}^{\mathrm{4}} }{\left({a}+{b}\right)^{\mathrm{2}} }−\frac{{ab}^{\mathrm{2}} }{{a}+{b}}=\mathrm{0} \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{{b}^{\mathrm{2}} }{{a}+{b}}\left(\mathrm{2}{a}+\frac{{b}^{\mathrm{2}} }{{a}+{b}}\right)−\sqrt{\left[\left(\mathrm{2}{ab}+\frac{{b}^{\mathrm{3}} }{{a}+{b}}\right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right]\frac{{b}^{\mathrm{4}} }{\left({a}+{b}\right)^{\mathrm{2}} }+\frac{\mathrm{4}{ab}^{\mathrm{2}} }{{a}+{b}}}\right\} \\ $$
Commented by mr W last updated on 02/Nov/24
Answered by ajfour last updated on 02/Nov/24
let center of circle be origin. y−(b^2 −y_c )=−((b^2 (x+b))/((a+b))) tan α=(b^2 /(a+b)) (r/(cos α))=b^2 −y_c −(b^3 /(a+b)) r=(((a+b))/( (√(b^4 +(a+b)^2 )))){b^2 −y_c −(b^3 /(a+b))} Eq. of parabola y=x^2 −y_c circle eq: x^2 +y^2 =r^2 (x^2 −y_c )^2 +x^2 =r^2 x^4 −(1−2y_c )x^2 +y_c ^2 −r^2 =0 △=0 ⇒ (1−2y_c )^2 =4(y_c ^2 −r^2 ) ⇒ 1−4y_c +4r^2 =0 since r=(((a+b))/( (√(b^4 +(a+b)^2 )))){b^2 −y_c −(b^3 /(a+b))} r=k(((ab^2 )/(a+b))−r^2 −(1/4)) r^2 +(r/k)−((ab^2 )/(a+b))+(1/4)=0 r=−(1/(2k))+(√((1/(4k^2 ))−(1/4)+((ab^2 )/(a+b)))) where k=(((a+b))/( (√(b^4 +(a+b)^2 ))))
$${let}\:{center}\:{of}\:{circle}\:{be}\:{origin}. \\ $$$${y}−\left({b}^{\mathrm{2}} −{y}_{{c}} \right)=−\frac{{b}^{\mathrm{2}} \left({x}+{b}\right)}{\left({a}+{b}\right)} \\ $$$$\mathrm{tan}\:\alpha=\frac{{b}^{\mathrm{2}} }{{a}+{b}} \\ $$$$\frac{{r}}{\mathrm{cos}\:\alpha}={b}^{\mathrm{2}} −{y}_{{c}} −\frac{{b}^{\mathrm{3}} }{{a}+{b}} \\ $$$${r}=\frac{\left({a}+{b}\right)}{\:\sqrt{{b}^{\mathrm{4}} +\left({a}+{b}\right)^{\mathrm{2}} }}\left\{{b}^{\mathrm{2}} −{y}_{{c}} −\frac{{b}^{\mathrm{3}} }{{a}+{b}}\right\} \\ $$$${Eq}.\:{of}\:{parabola} \\ $$$${y}={x}^{\mathrm{2}} −{y}_{{c}} \\ $$$${circle}\:{eq}:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} −{y}_{{c}} \right)^{\mathrm{2}} +{x}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\left(\mathrm{1}−\mathrm{2}{y}_{{c}} \right){x}^{\mathrm{2}} +{y}_{{c}} ^{\mathrm{2}} −{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\bigtriangleup=\mathrm{0}\:\:\Rightarrow\:\:\left(\mathrm{1}−\mathrm{2}{y}_{{c}} \right)^{\mathrm{2}} =\mathrm{4}\left({y}_{{c}} ^{\mathrm{2}} −{r}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:\mathrm{1}−\mathrm{4}{y}_{{c}} +\mathrm{4}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${since} \\ $$$${r}=\frac{\left({a}+{b}\right)}{\:\sqrt{{b}^{\mathrm{4}} +\left({a}+{b}\right)^{\mathrm{2}} }}\left\{{b}^{\mathrm{2}} −{y}_{{c}} −\frac{{b}^{\mathrm{3}} }{{a}+{b}}\right\} \\ $$$${r}={k}\left(\frac{{ab}^{\mathrm{2}} }{{a}+{b}}−{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${r}^{\mathrm{2}} +\frac{{r}}{{k}}−\frac{{ab}^{\mathrm{2}} }{{a}+{b}}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$${r}=−\frac{\mathrm{1}}{\mathrm{2}{k}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}{k}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}+\frac{{ab}^{\mathrm{2}} }{{a}+{b}}} \\ $$$${where}\:\:\:{k}=\frac{\left({a}+{b}\right)}{\:\sqrt{{b}^{\mathrm{4}} +\left({a}+{b}\right)^{\mathrm{2}} }} \\ $$
Commented by ajfour last updated on 02/Nov/24
For a=4, b=2 (1/(2k))=((√(16+36))/(12))=((√(13))/6) r=−((√(13))/6)+(√((13−9+96)/(36))) r =((10−(√(13)))/6) ≈ 1.06574 y_c =(1/4)+((113−20(√(13)))/(36)) =((61−10(√(13)))/(18)) y_c ≈ 1.3858
$${For}\:{a}=\mathrm{4},\:{b}=\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{k}}=\frac{\sqrt{\mathrm{16}+\mathrm{36}}}{\mathrm{12}}=\frac{\sqrt{\mathrm{13}}}{\mathrm{6}} \\ $$$${r}=−\frac{\sqrt{\mathrm{13}}}{\mathrm{6}}+\sqrt{\frac{\mathrm{13}−\mathrm{9}+\mathrm{96}}{\mathrm{36}}} \\ $$$$\:\:\boldsymbol{{r}}\:=\frac{\mathrm{10}−\sqrt{\mathrm{13}}}{\mathrm{6}}\:\approx\:\mathrm{1}.\mathrm{06574} \\ $$$${y}_{{c}} =\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{113}−\mathrm{20}\sqrt{\mathrm{13}}}{\mathrm{36}}\:=\frac{\mathrm{61}−\mathrm{10}\sqrt{\mathrm{13}}}{\mathrm{18}} \\ $$$$\:\:\boldsymbol{{y}}_{\boldsymbol{{c}}} \approx\:\mathrm{1}.\mathrm{3858} \\ $$
Commented by ajfour last updated on 02/Nov/24

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