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Question Number 46012 by peter frank last updated on 19/Oct/18

the normal at any point of hyperbola meets the axes at E,F.find the locus of the midpoint of EF.

$$\mathrm{the}\:\mathrm{normal}\:\mathrm{at}\:\mathrm{any}\:\mathrm{point} \\ $$$$\mathrm{of}\:\mathrm{hyperbola}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{axes} \\ $$$$\mathrm{at}\:\mathrm{E},\mathrm{F}.\mathrm{find}\:\mathrm{the}\:\mathrm{locus}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{EF}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

(x^2 /a^2 )−(y^2 /b^2 )=1 eqn of hyperbola (asecθ,btanθ) iz a point on hyperbola slope at (asecθ,btanθ)... (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 ))))=((xb^2 )/(ya^2 )) so m=((asecθ×b^2 )/(btanθ×a^2 ))=((bsecθ)/(atanθ)) slope of normal m_1 so m×m_1 =−1 m_1 =((−1)/m)=((−atanθ)/(bsecθ)) eqn normal at (asecθ,btanθ)is (y−btanθ)=((−atanθ)/(bsecθ))(x−asecθ) ybsecθ−b^2 tanθsecθ=−xatanθ+a^2 tanθsecθ xatanθ+ybsecθ=(a^2 +b^2 )tanθsecθ (x/(((a^2 +b^2 )tanθsecθ)/(atanθ)))+(y/(((a^2 +b^2 )tanθsecθ)/(bsecθ)))=1 so point E(((a^2 +b^2 )/a)secθ,0) point F(0,((a^2 +b^2 )/b)tanθ) mid point of E,F (α,β) α=((a^2 +b^2 )/(2a))secθ β=((a^2 +b^2 )/(2b))tanθ we know sec^2 θ−tan^2 θ=1 (((2aα)/(a^2 +b^2 )))^2 −(((2bβ)/(a^2 +b^2 )))^2 =1 4a^2 α^2 −4b^2 β^2 =(a^2 +b^2 )^2 hence locus is 4a^2 x^2 −4b^2 y^2 =(a^2 +b^2 )^2 pls check

$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:{eqn}\:{of}\:{hyperbola}\:\: \\ $$$$\left({asec}\theta,{btan}\theta\right)\:{iz}\:{a}\:{point}\:{on}\:{hyperbola} \\ $$$${slope}\:{at}\:\left({asec}\theta,{btan}\theta\right)... \\ $$$$\frac{{d}}{{dx}}\left(\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\frac{\mathrm{2}{x}}{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{y}}{{b}^{\mathrm{2}} }×\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=\frac{\left(\frac{−\mathrm{2}{x}}{{a}^{\mathrm{2}} }\right)}{\left(−\frac{\mathrm{2}{y}}{{b}^{\mathrm{2}} }\right)}=\frac{{xb}^{\mathrm{2}} }{{ya}^{\mathrm{2}} }\:\:{so}\:{m}=\frac{{asec}\theta×{b}^{\mathrm{2}} }{{btan}\theta×{a}^{\mathrm{2}} }=\frac{{bsec}\theta}{{atan}\theta} \\ $$$${slope}\:{of}\:{normal}\:{m}_{\mathrm{1}} \:\:\:\:{so}\:{m}×{m}_{\mathrm{1}} =−\mathrm{1} \\ $$$${m}_{\mathrm{1}} =\frac{−\mathrm{1}}{{m}}=\frac{−{atan}\theta}{{bsec}\theta} \\ $$$${eqn}\:{normal}\:{at}\:\left({asec}\theta,{btan}\theta\right){is} \\ $$$$\left({y}−{btan}\theta\right)=\frac{−{atan}\theta}{{bsec}\theta}\left({x}−{asec}\theta\right) \\ $$$${ybsec}\theta−{b}^{\mathrm{2}} {tan}\theta{sec}\theta=−{xatan}\theta+{a}^{\mathrm{2}} {tan}\theta{sec}\theta \\ $$$${xatan}\theta+{ybsec}\theta=\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){tan}\theta{sec}\theta \\ $$$$\frac{{x}}{\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){tan}\theta{sec}\theta}{{atan}\theta}}+\frac{{y}}{\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){tan}\theta{sec}\theta}{{bsec}\theta}}=\mathrm{1} \\ $$$${so}\:{point}\:{E}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}}{sec}\theta,\mathrm{0}\right)\:\:{point}\:{F}\left(\mathrm{0},\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}}{tan}\theta\right) \\ $$$${mid}\:{point}\:{of}\:{E},{F}\:\:\:\left(\alpha,\beta\right) \\ $$$$\alpha=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}{a}}{sec}\theta\:\:\:\:\beta=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}{b}}{tan}\theta \\ $$$${we}\:{know}\:{sec}^{\mathrm{2}} \theta−{tan}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\left(\frac{\mathrm{2}{a}\alpha}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{\mathrm{2}} −\left(\frac{\mathrm{2}{b}\beta}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} \alpha^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} \beta^{\mathrm{2}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${hence}\:{locus}\:{is} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} {y}^{\mathrm{2}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:\:\:\:{pls}\:{check} \\ $$$$ \\ $$

Commented by peter frank last updated on 20/Oct/18

yes sir you right thanks

$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{you}\:\mathrm{right}\:\mathrm{thanks} \\ $$

Answered by ajfour last updated on 20/Oct/18

let the midpoint be (h,k) eq. of normal is: y= −((kx)/h)+2k ....(i) hyperbola: (x^2 /a^2 )−(y^2 /b^2 ) = 1 ⇒ (dy/dx)∣_(x_1 ,y_1 ) =(b^2 /a^2 )((x_1 /y_1 )) eq. of normal: y = −((a^2 y_1 )/(b^2 x_1 ))(x−x_1 )+y_1 ⇒ y = −(((a^2 y_1 )/(b^2 x_1 )))x+y_1 (1+(a^2 /b^2 )) ..(ii) comparing (i) & (ii) (k/h) = ((a^2 y_1 )/(b^2 x_1 )) & 2k = y_1 (1+(a^2 /b^2 )) also (x_1 ^2 /a^2 ) = 1+(y_1 ^2 /b^2 ) ⇒ (k^2 /h^2 )= (a^4 /b^4 )×(y_1 ^2 /(a^2 (1+(y_1 ^2 /b^2 )))) (k^2 /h^2 ) = (a^2 /b^2 )×(1/(((b^2 /y_1 ^2 )+1))) ⇒ (k^2 /h^2 )= (a^2 /b^2 )×(1/([(((a^2 +b^2 )^2 )/(4b^2 k^2 ))+1])) ⇒ (k^2 /h^2 )= (a^2 /b^2 )×((4b^2 k^2 )/([(a^2 +b^2 )^2 +4b^2 k^2 ])) ⇒ 4b^2 k^2 +(a^2 +b^2 )^2 = 4a^2 h^2 hence required locus is 4(a^2 x^2 −b^2 y^2 )=(a^2 +b^2 )^2 .

$${let}\:{the}\:{midpoint}\:{be}\:\left({h},{k}\right) \\ $$$${eq}.\:{of}\:{normal}\:{is}: \\ $$$$\:\:\:\:\:\:\:{y}=\:−\frac{{kx}}{{h}}+\mathrm{2}{k}\:\:\:\:\:\:\:\:....\left({i}\right) \\ $$$${hyperbola}:\:\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\:\:\frac{{dy}}{{dx}}\mid_{{x}_{\mathrm{1}} ,{y}_{\mathrm{1}} } =\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\left(\frac{{x}_{\mathrm{1}} }{{y}_{\mathrm{1}} }\right)\: \\ $$$${eq}.\:{of}\:{normal}: \\ $$$$\:\:\:\:{y}\:=\:−\frac{{a}^{\mathrm{2}} {y}_{\mathrm{1}} }{{b}^{\mathrm{2}} {x}_{\mathrm{1}} }\left({x}−{x}_{\mathrm{1}} \right)+{y}_{\mathrm{1}} \\ $$$$\Rightarrow\:\:{y}\:=\:−\left(\frac{{a}^{\mathrm{2}} {y}_{\mathrm{1}} }{{b}^{\mathrm{2}} {x}_{\mathrm{1}} }\right){x}+{y}_{\mathrm{1}} \left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)\:..\left({ii}\right) \\ $$$${comparing}\:\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\:\:\:\:\:\frac{{k}}{{h}}\:=\:\frac{{a}^{\mathrm{2}} {y}_{\mathrm{1}} }{{b}^{\mathrm{2}} {x}_{\mathrm{1}} }\:\:\:\&\:\:\:\mathrm{2}{k}\:=\:{y}_{\mathrm{1}} \left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right) \\ $$$${also}\:\:\:\:\frac{{x}_{\mathrm{1}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:=\:\mathrm{1}+\frac{{y}_{\mathrm{1}} ^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\frac{{k}^{\mathrm{2}} }{{h}^{\mathrm{2}} }=\:\frac{{a}^{\mathrm{4}} }{{b}^{\mathrm{4}} }×\frac{{y}_{\mathrm{1}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} \left(\mathrm{1}+\frac{{y}_{\mathrm{1}} ^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)} \\ $$$$\:\:\:\:\frac{{k}^{\mathrm{2}} }{{h}^{\mathrm{2}} }\:=\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }×\frac{\mathrm{1}}{\left(\frac{{b}^{\mathrm{2}} }{{y}_{\mathrm{1}} ^{\mathrm{2}} }+\mathrm{1}\right)} \\ $$$$\Rightarrow\:\:\frac{{k}^{\mathrm{2}} }{{h}^{\mathrm{2}} }=\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }×\frac{\mathrm{1}}{\left[\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} {k}^{\mathrm{2}} }+\mathrm{1}\right]} \\ $$$$\Rightarrow\:\:\frac{{k}^{\mathrm{2}} }{{h}^{\mathrm{2}} }=\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }×\frac{\mathrm{4}{b}^{\mathrm{2}} {k}^{\mathrm{2}} }{\left[\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} {k}^{\mathrm{2}} \right]} \\ $$$$\Rightarrow\:\mathrm{4}{b}^{\mathrm{2}} {k}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} =\:\mathrm{4}{a}^{\mathrm{2}} {h}^{\mathrm{2}} \\ $$$$\:\:{hence}\:{required}\:{locus}\:{is} \\ $$$$\:\:\:\mathrm{4}\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} −{b}^{\mathrm{2}} {y}^{\mathrm{2}} \right)=\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:. \\ $$

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