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Question Number 164728 by mr W last updated on 21/Jan/22

Commented by mr W last updated on 21/Jan/22

two points P and Q on the parabola y=(x^2 /a) with ∣PQ∣=l. the normal lines at P and Q intersect at point S. find the locus of S.

$${two}\:{points}\:{P}\:{and}\:{Q}\:{on}\:{the}\:{parabola} \\ $$$${y}=\frac{{x}^{\mathrm{2}} }{{a}}\:{with}\:\mid{PQ}\mid={l}.\:{the}\:{normal}\:{lines} \\ $$$${at}\:{P}\:{and}\:{Q}\:{intersect}\:{at}\:{point}\:{S}. \\ $$$${find}\:{the}\:{locus}\:{of}\:{S}. \\ $$

Answered by ajfour last updated on 21/Jan/22

S(−h,k) Normal from S y=−((a/(2p)))x+c P(p,(p^2 /a)) lies on it (p^2 /a)=−(a/2)+c (p_2 −p_1 )^2 +(((p_2 ^2 −p_1 ^2 )^2 )/a^3 )=L^2 ...(I) k=((ah)/(2p))+(p^2 /a)+(a/2) ⇒ 2p^3 +a(a−2k)p+ah=0 we get p_1 ,p_2 , p_3 in terms of h, k,a . substituting in ..(I) gives required the three locii, for three choices of pairs from among p_1 , p_2 , p_3 .

$${S}\left(−{h},{k}\right) \\ $$$${Normal}\:\:{from}\:{S} \\ $$$${y}=−\left(\frac{{a}}{\mathrm{2}{p}}\right){x}+{c}\:\:\:\:\:\:\:{P}\left({p},\frac{{p}^{\mathrm{2}} }{{a}}\right)\:\:{lies}\:{on}\:{it} \\ $$$$\frac{{p}^{\mathrm{2}} }{{a}}=−\frac{{a}}{\mathrm{2}}+{c} \\ $$$$\left({p}_{\mathrm{2}} −{p}_{\mathrm{1}} \right)^{\mathrm{2}} +\frac{\left({p}_{\mathrm{2}} ^{\mathrm{2}} −{p}_{\mathrm{1}} ^{\mathrm{2}} \right)^{\mathrm{2}} }{{a}^{\mathrm{3}} }={L}^{\mathrm{2}} \:\:\:\:...\left({I}\right) \\ $$$${k}=\frac{{ah}}{\mathrm{2}{p}}+\frac{{p}^{\mathrm{2}} }{{a}}+\frac{{a}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{2}{p}^{\mathrm{3}} +{a}\left({a}−\mathrm{2}{k}\right){p}+{ah}=\mathrm{0} \\ $$$${we}\:{get}\:{p}_{\mathrm{1}} ,{p}_{\mathrm{2}} ,\:{p}_{\mathrm{3}} \:\:{in}\:{terms}\:{of} \\ $$$$\:\:{h},\:{k},{a}\:. \\ $$$${substituting}\:{in}\:..\left({I}\right)\:{gives}\:{required} \\ $$$${the}\:{three}\:{locii},\:{for}\:{three}\:{choices}\:{of} \\ $$$${pairs}\:{from}\:{among}\:{p}_{\mathrm{1}} ,\:{p}_{\mathrm{2}} ,\:{p}_{\mathrm{3}} . \\ $$$$\:\: \\ $$

Commented by mr W last updated on 21/Jan/22

thanks for trying sir! i′ll also give a try.

$${thanks}\:{for}\:{trying}\:{sir}!\:{i}'{ll}\:{also}\:{give}\:{a} \\ $$$${try}. \\ $$

Answered by mr W last updated on 21/Jan/22

Commented by mr W last updated on 23/Jan/22

locus in shape of a German Brezel...

$${locus}\:{in}\:{shape}\:{of}\:{a}\:{German}\:{Brezel}... \\ $$

Commented by mr W last updated on 22/Jan/22

Commented by mr W last updated on 21/Jan/22

say P(p, (p^2 /a)), Q(q, (q^2 /a)) tan φ=((2p)/a) tan ϕ=((2q)/a) (q−p)^2 +((q^2 /a)−(p^2 /a))^2 =l^2 (q−p)^2 [1+(((p+q)^2 )/a^2 )]=l^2 [(q+p)^2 −4pq][1+(((p+q)^2 )/a^2 )]=l^2 let u=p+q, v=pq (u^2 −4v)(1+(u^2 /a^2 ))=l^2 ⇒v=(1/4)(u^2 −((a^2 l^2 )/(a^2 +u^2 ))) eqn. of PS: y=(p^2 /a)−(a/(2p))(x−p) y=(p^2 /a)+(a/2)−((ax)/(2p)) eqn. of QS: y=(q^2 /a)+(a/2)−((ax)/(2q)) say intersection point S(x_s ,y_s ) (p^2 /a)+(a/2)−((ax_s )/(2p))=(q^2 /a)+(a/2)−((ax_s )/(2q)) (p^2 /a)−((ax_s )/(2p))=(q^2 /a)−((ax_s )/(2q)) x_s =−((2(q+p)pq)/a^2 ) ⇒x_s =−((2uv)/a^2 ) 2y_s =((p^2 +q^2 )/a)+a−ax_s ((1/(2p))+(1/(2q))) y_s =(a/2)+((p^2 +q^2 )/(2a))−((a(p+q)x_s )/(4pq)) y_s =(a/2)+(((p+q)^2 −pq)/a) ⇒y_s =(a/2)+((u^2 −v)/a) taking u as parameter we get the eqn. of locus of point S: { ((x=−((2uv)/a^2 ))),((y=(a/2)+((u^2 −v)/a))) :} with v=(1/4)(u^2 −((a^2 l^2 )/(a^2 +u^2 ))) with u given, p, q are roots of z^2 −uz+(1/4)(u^2 −((a^2 l^2 )/(a^2 +u^2 )))=0 ⇒p,q=(1/2)(u∓((al)/( (√(a^2 +u^2 )))))

$${say}\:{P}\left({p},\:\frac{{p}^{\mathrm{2}} }{{a}}\right),\:{Q}\left({q},\:\frac{{q}^{\mathrm{2}} }{{a}}\right) \\ $$$$\mathrm{tan}\:\phi=\frac{\mathrm{2}{p}}{{a}} \\ $$$$\mathrm{tan}\:\varphi=\frac{\mathrm{2}{q}}{{a}} \\ $$$$\left({q}−{p}\right)^{\mathrm{2}} +\left(\frac{{q}^{\mathrm{2}} }{{a}}−\frac{{p}^{\mathrm{2}} }{{a}}\right)^{\mathrm{2}} ={l}^{\mathrm{2}} \\ $$$$\left({q}−{p}\right)^{\mathrm{2}} \left[\mathrm{1}+\frac{\left({p}+{q}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]={l}^{\mathrm{2}} \\ $$$$\left[\left({q}+{p}\right)^{\mathrm{2}} −\mathrm{4}{pq}\right]\left[\mathrm{1}+\frac{\left({p}+{q}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]={l}^{\mathrm{2}} \\ $$$${let}\:{u}={p}+{q},\:{v}={pq} \\ $$$$\left({u}^{\mathrm{2}} −\mathrm{4}{v}\right)\left(\mathrm{1}+\frac{{u}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)={l}^{\mathrm{2}} \\ $$$$\Rightarrow{v}=\frac{\mathrm{1}}{\mathrm{4}}\left({u}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} {l}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{u}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$${eqn}.\:{of}\:{PS}: \\ $$$${y}=\frac{{p}^{\mathrm{2}} }{{a}}−\frac{{a}}{\mathrm{2}{p}}\left({x}−{p}\right) \\ $$$${y}=\frac{{p}^{\mathrm{2}} }{{a}}+\frac{{a}}{\mathrm{2}}−\frac{{ax}}{\mathrm{2}{p}} \\ $$$${eqn}.\:{of}\:{QS}: \\ $$$${y}=\frac{{q}^{\mathrm{2}} }{{a}}+\frac{{a}}{\mathrm{2}}−\frac{{ax}}{\mathrm{2}{q}} \\ $$$${say}\:{intersection}\:{point}\:{S}\left({x}_{{s}} ,{y}_{{s}} \right) \\ $$$$\frac{{p}^{\mathrm{2}} }{{a}}+\frac{{a}}{\mathrm{2}}−\frac{{ax}_{{s}} }{\mathrm{2}{p}}=\frac{{q}^{\mathrm{2}} }{{a}}+\frac{{a}}{\mathrm{2}}−\frac{{ax}_{{s}} }{\mathrm{2}{q}} \\ $$$$\frac{{p}^{\mathrm{2}} }{{a}}−\frac{{ax}_{{s}} }{\mathrm{2}{p}}=\frac{{q}^{\mathrm{2}} }{{a}}−\frac{{ax}_{{s}} }{\mathrm{2}{q}} \\ $$$${x}_{{s}} =−\frac{\mathrm{2}\left({q}+{p}\right){pq}}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}_{{s}} =−\frac{\mathrm{2}{uv}}{{a}^{\mathrm{2}} } \\ $$$$\mathrm{2}{y}_{{s}} =\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{{a}}+{a}−{ax}_{{s}} \left(\frac{\mathrm{1}}{\mathrm{2}{p}}+\frac{\mathrm{1}}{\mathrm{2}{q}}\right) \\ $$$${y}_{{s}} =\frac{{a}}{\mathrm{2}}+\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{\mathrm{2}{a}}−\frac{{a}\left({p}+{q}\right){x}_{{s}} }{\mathrm{4}{pq}} \\ $$$${y}_{{s}} =\frac{{a}}{\mathrm{2}}+\frac{\left({p}+{q}\right)^{\mathrm{2}} −{pq}}{{a}} \\ $$$$\Rightarrow{y}_{{s}} =\frac{{a}}{\mathrm{2}}+\frac{{u}^{\mathrm{2}} −{v}}{{a}} \\ $$$${taking}\:{u}\:{as}\:{parameter}\:{we}\:{get}\:{the}\:{eqn}. \\ $$$${of}\:{locus}\:{of}\:{point}\:{S}: \\ $$$$\begin{cases}{{x}=−\frac{\mathrm{2}{uv}}{{a}^{\mathrm{2}} }}\\{{y}=\frac{{a}}{\mathrm{2}}+\frac{{u}^{\mathrm{2}} −{v}}{{a}}}\end{cases}\:{with}\:{v}=\frac{\mathrm{1}}{\mathrm{4}}\left({u}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} {l}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{u}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$${with}\:{u}\:{given},\:{p},\:{q}\:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{2}} −{uz}+\frac{\mathrm{1}}{\mathrm{4}}\left({u}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} {l}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{u}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\Rightarrow{p},{q}=\frac{\mathrm{1}}{\mathrm{2}}\left({u}\mp\frac{{al}}{\:\sqrt{{a}^{\mathrm{2}} +{u}^{\mathrm{2}} }}\right) \\ $$

Commented by mr W last updated on 21/Jan/22

Commented by mr W last updated on 21/Jan/22

Commented by ajfour last updated on 21/Jan/22

great workings sir, i will try n follow..

$${great}\:{workings}\:{sir},\:{i}\:{will}\:{try}\:{n} \\ $$$${follow}.. \\ $$

Commented by Tawa11 last updated on 21/Jan/22

Weldone sir

$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

Answered by ajfour last updated on 23/Jan/22

S(h,k) ((q−h)/(k−(q^2 /a)))=((2p)/a) ⇒ ak−q^2 =(a^2 /2)(((q−h)/p)) ((p−h)/(k−(p^2 /a)))=((2q)/a) ⇒ ak−p^2 =(a^2 /2)(((p−h)/q)) subtracting ⇒ q^2 −p^2 =(a^2 /2)((p/q)−(q/p))+((a^2 h)/2)((1/p)−(1/q)) ⇒ q+p=((a^2 h)/(2pq))−((a^2 (q+p))/(2pq)) ⇒ q+p=((a^2 h)/(a^2 +2pq)) say q+p=s, pq=m ⇒ s=((a^2 h)/(a^2 +2m)) ...(i) Adding 2ak−(p^2 +q^2 )=(a^2 /2)((p/q)+(q/p))−((a^2 h)/2)((1/p)+(1/q)) ⇒ 2ak−s^2 +2m=(a^2 /2)(((s^2 −2m)/m)−((hs)/m)) .....(ii) And s^2 +((s^2 (s^2 −4m))/a^2 )=L^2 ...(iii) ⇒ using (i) and (iii) 1+((s^2 −4m)/a^2 )=(L^2 /h^2 )(1+(m/a^2 )) ⇒ m((L^2 /(a^2 h^2 ))+(4/a^2 ))=1−(L^2 /h^2 ) ⇒ m=((a^2 (1−(L^2 /h^2 )))/((4+(L^2 /h^2 )))) (1/s)=(1/h)+((2m)/(a^2 h))=(1/h)+(2/h)(((1−(L^2 /h^2 ))/(4+(L^2 /h^2 )))) Now substituting for m, s in (ii) ((2ak)/s^2 )+1+((2m)/s^2 )=(a^2 /2)((1/m)−(2/s^2 )−(h/(ms))) {2ak+2a^2 (((h^2 −L^2 )/(4h^2 +L^2 )))+a^2 }{(1/h)+(2/h)(((h^2 −L^2 )/(4h^2 +L^2 )))}^2 +2 =0 And to make it compact say (h/L)=X , (k/L)=Y ((2a)/L)(Y+((X^2 −1)/(4X^2 +1))+(a/L)){1+2(((X^2 −1)/(4X^2 +1)))}^2 +2X^2 =0 or simply (X/(((1/2)+((X^2 −1)/(4X^2 +1)))))=±(((2a)/L))^(1/2) (√(((1−X^2 )/(4X^2 +1))−Y−(a/L))) say for a=1, L=2 (x/({1+((x^2 −4)/(2(x^2 +1)))}))=±(√(((4−x^2 )/(4(x^2 +1)))−(y/2)−(1/2))) ⇒ y=((4−x^2 )/(2(x^2 +1)))−((16(x^2 +1)^2 )/((3x^2 −2)^2 ))−1

$${S}\left({h},{k}\right) \\ $$$$\frac{{q}−{h}}{{k}−\frac{{q}^{\mathrm{2}} }{{a}}}=\frac{\mathrm{2}{p}}{{a}}\:\:\:\Rightarrow\:\:{ak}−{q}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{{q}−{h}}{{p}}\right) \\ $$$$\:\:\frac{{p}−{h}}{{k}−\frac{{p}^{\mathrm{2}} }{{a}}}=\frac{\mathrm{2}{q}}{{a}}\:\:\Rightarrow\:\:{ak}−{p}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{{p}−{h}}{{q}}\right) \\ $$$${subtracting} \\ $$$$\Rightarrow\:{q}^{\mathrm{2}} −{p}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{{p}}{{q}}−\frac{{q}}{{p}}\right)+\frac{{a}^{\mathrm{2}} {h}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{p}}−\frac{\mathrm{1}}{{q}}\right) \\ $$$$\Rightarrow\:\:{q}+{p}=\frac{{a}^{\mathrm{2}} {h}}{\mathrm{2}{pq}}−\frac{{a}^{\mathrm{2}} \left({q}+{p}\right)}{\mathrm{2}{pq}} \\ $$$$\Rightarrow\:\:\:{q}+{p}=\frac{{a}^{\mathrm{2}} {h}}{{a}^{\mathrm{2}} +\mathrm{2}{pq}} \\ $$$${say}\:\:{q}+{p}={s},\:\:\:{pq}={m}\:\: \\ $$$$\Rightarrow\:\:\:{s}=\frac{{a}^{\mathrm{2}} {h}}{{a}^{\mathrm{2}} +\mathrm{2}{m}}\:\:\:\:...\left({i}\right) \\ $$$$\:\:{Adding} \\ $$$$\mathrm{2}{ak}−\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{{p}}{{q}}+\frac{{q}}{{p}}\right)−\frac{{a}^{\mathrm{2}} {h}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}\right) \\ $$$$\Rightarrow\:\:\mathrm{2}{ak}−{s}^{\mathrm{2}} +\mathrm{2}{m}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{{s}^{\mathrm{2}} −\mathrm{2}{m}}{{m}}−\frac{{hs}}{{m}}\right) \\ $$$$\:\:\:.....\left({ii}\right) \\ $$$${And}\:\:\:{s}^{\mathrm{2}} +\frac{{s}^{\mathrm{2}} \left({s}^{\mathrm{2}} −\mathrm{4}{m}\right)}{{a}^{\mathrm{2}} }={L}^{\mathrm{2}} \:\:\:...\left({iii}\right) \\ $$$$\Rightarrow\:\:{using}\:\left({i}\right)\:{and}\:\left({iii}\right) \\ $$$$\mathrm{1}+\frac{{s}^{\mathrm{2}} −\mathrm{4}{m}}{{a}^{\mathrm{2}} }=\frac{{L}^{\mathrm{2}} }{{h}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{m}}{{a}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\:\:{m}\left(\frac{{L}^{\mathrm{2}} }{{a}^{\mathrm{2}} {h}^{\mathrm{2}} }+\frac{\mathrm{4}}{{a}^{\mathrm{2}} }\right)=\mathrm{1}−\frac{{L}^{\mathrm{2}} }{{h}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:{m}=\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−\frac{{L}^{\mathrm{2}} }{{h}^{\mathrm{2}} }\right)}{\left(\mathrm{4}+\frac{{L}^{\mathrm{2}} }{{h}^{\mathrm{2}} }\right)} \\ $$$$\:\:\:\frac{\mathrm{1}}{{s}}=\frac{\mathrm{1}}{{h}}+\frac{\mathrm{2}{m}}{{a}^{\mathrm{2}} {h}}=\frac{\mathrm{1}}{{h}}+\frac{\mathrm{2}}{{h}}\left(\frac{\mathrm{1}−\frac{{L}^{\mathrm{2}} }{{h}^{\mathrm{2}} }}{\mathrm{4}+\frac{{L}^{\mathrm{2}} }{{h}^{\mathrm{2}} }}\right) \\ $$$${Now}\:\:{substituting}\:{for}\:{m},\:{s}\:{in}\:\left({ii}\right) \\ $$$$\frac{\mathrm{2}{ak}}{{s}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{2}{m}}{{s}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\mathrm{1}}{{m}}−\frac{\mathrm{2}}{{s}^{\mathrm{2}} }−\frac{{h}}{{ms}}\right) \\ $$$$\left\{\mathrm{2}{ak}+\mathrm{2}{a}^{\mathrm{2}} \left(\frac{{h}^{\mathrm{2}} −{L}^{\mathrm{2}} }{\mathrm{4}{h}^{\mathrm{2}} +{L}^{\mathrm{2}} }\right)+{a}^{\mathrm{2}} \right\}\left\{\frac{\mathrm{1}}{{h}}+\frac{\mathrm{2}}{{h}}\left(\frac{{h}^{\mathrm{2}} −{L}^{\mathrm{2}} }{\mathrm{4}{h}^{\mathrm{2}} +{L}^{\mathrm{2}} }\right)\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:+\mathrm{2}\:=\mathrm{0} \\ $$$${And}\:{to}\:{make}\:{it}\:{compact} \\ $$$${say}\:\:\frac{{h}}{{L}}={X}\:,\:\frac{{k}}{{L}}={Y} \\ $$$$\frac{\mathrm{2}{a}}{{L}}\left({Y}+\frac{{X}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}{X}^{\mathrm{2}} +\mathrm{1}}+\frac{{a}}{{L}}\right)\left\{\mathrm{1}+\mathrm{2}\left(\frac{{X}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}{X}^{\mathrm{2}} +\mathrm{1}}\right)\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{X}^{\mathrm{2}} =\mathrm{0} \\ $$$${or}\:{simply} \\ $$$$\:\:\frac{{X}}{\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{{X}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}{X}^{\mathrm{2}} +\mathrm{1}}\right)}=\pm\left(\frac{\mathrm{2}{a}}{{L}}\right)^{\mathrm{1}/\mathrm{2}} \sqrt{\frac{\mathrm{1}−{X}^{\mathrm{2}} }{\mathrm{4}{X}^{\mathrm{2}} +\mathrm{1}}−{Y}−\frac{{a}}{{L}}} \\ $$$${say}\:\:\:{for}\:\:{a}=\mathrm{1},\:{L}=\mathrm{2} \\ $$$$\:\frac{{x}}{\left\{\mathrm{1}+\frac{{x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\right\}}=\pm\sqrt{\frac{\mathrm{4}−{x}^{\mathrm{2}} }{\mathrm{4}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{{y}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\:\:{y}=\frac{\mathrm{4}−{x}^{\mathrm{2}} }{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{16}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} }−\mathrm{1} \\ $$

Commented by ajfour last updated on 23/Jan/22

some little error ...i shall edit today..

$${some}\:{little}\:{error}\:...{i}\:{shall}\:{edit} \\ $$$${today}.. \\ $$

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