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a-Normal-to-any-point-on-the-hyperbola-XY-C-meet-the-x-axis-at-A-and-tangents-meets-the-y-axis-at-B-find-the-locus-of-the-mid-point-of-AB-b-find-the-equation-of-assymptotes-of-i-x-2-4-y-2-5-1

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Question Number 50979 by peter frank last updated on 23/Dec/18
a)Normal to any point on the hyperbola XY=C meet the x−axis at A and tangents meets the y−axis at B.find the locus of the mid point of AB b)find the equation of assymptotes of (i)(x^2 /4)−(y^2 /5)=1 (ii)(((x−1)^2 )/(16))−(((y−3)^2 )/9)=1
$$\left.{a}\right){Normal}\:{to}\:{any}\:{point}\:{on} \\ $$$${the}\:{hyperbola}\:{XY}={C} \\ $$$${meet}\:{the}\:{x}−{axis}\:{at}\:{A} \\ $$$${and}\:{tangents}\:{meets} \\ $$$${the}\:{y}−{axis}\:{at}\:{B}.{find}\:{the} \\ $$$${locus}\:{of}\:{the}\:{mid}\:{point}\:{of}\:{AB} \\ $$$$\left.{b}\right){find}\:\:{the}\:{equation}\:{of}\: \\ $$$${assymptotes}\:{of} \\ $$$$\left({i}\right)\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{{y}^{\mathrm{2}} }{\mathrm{5}}=\mathrm{1} \\ $$$$\left({ii}\right)\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{16}}−\frac{\left({y}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1} \\ $$
Commented by ggururajguru0219@gmail.com last updated on 23/Dec/18
vg5n
$${vg}\mathrm{5}{n} \\ $$
Answered by mr W last updated on 23/Dec/18
(a) xy=c y′=−(c/x^2 ) P(h,(c/h)) y′=−(c/h^2 ) tangent: y=(c/h)−(c/h^2 )(x−h) B(0,y_B ) ⇒y_B =((2c)/h) normal: y=(c/h)+(h^2 /c)(x−h) A(x_A ,0) 0=(c/h)+(h^2 /c)(x_A −h) ⇒x_A =h−(c^2 /h^3 ) mid point of AB: (p,q) q=(y_B /2)=(1/2)×((2c)/h)=(c/h) ⇒h=(c/q) p=(x_A /2)=(1/2)(h−(c^2 /h^3 ))=(1/2)((c/q)−(q^3 /c)) 2cpq=c^2 −q^4 ⇒2cxy=c^2 −y^4
$$\left({a}\right) \\ $$$${xy}={c} \\ $$$${y}'=−\frac{{c}}{{x}^{\mathrm{2}} } \\ $$$${P}\left({h},\frac{{c}}{{h}}\right) \\ $$$${y}'=−\frac{{c}}{{h}^{\mathrm{2}} } \\ $$$${tangent}: \\ $$$${y}=\frac{{c}}{{h}}−\frac{{c}}{{h}^{\mathrm{2}} }\left({x}−{h}\right) \\ $$$${B}\left(\mathrm{0},{y}_{{B}} \right) \\ $$$$\Rightarrow{y}_{{B}} =\frac{\mathrm{2}{c}}{{h}} \\ $$$${normal}: \\ $$$${y}=\frac{{c}}{{h}}+\frac{{h}^{\mathrm{2}} }{{c}}\left({x}−{h}\right) \\ $$$${A}\left({x}_{{A}} ,\mathrm{0}\right) \\ $$$$\mathrm{0}=\frac{{c}}{{h}}+\frac{{h}^{\mathrm{2}} }{{c}}\left({x}_{{A}} −{h}\right) \\ $$$$\Rightarrow{x}_{{A}} ={h}−\frac{{c}^{\mathrm{2}} }{{h}^{\mathrm{3}} } \\ $$$${mid}\:{point}\:{of}\:{AB}:\:\left({p},{q}\right) \\ $$$${q}=\frac{{y}_{{B}} }{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{c}}{{h}}=\frac{{c}}{{h}} \\ $$$$\Rightarrow{h}=\frac{{c}}{{q}} \\ $$$${p}=\frac{{x}_{{A}} }{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left({h}−\frac{{c}^{\mathrm{2}} }{{h}^{\mathrm{3}} }\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{c}}{{q}}−\frac{{q}^{\mathrm{3}} }{{c}}\right) \\ $$$$\mathrm{2}{cpq}={c}^{\mathrm{2}} −{q}^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{2}{cxy}={c}^{\mathrm{2}} −{y}^{\mathrm{4}} \\ $$
Commented by mr W last updated on 23/Dec/18
Commented by peter frank last updated on 23/Dec/18
thank you
$${thank}\:{you} \\ $$
Commented by peter frank last updated on 23/Dec/18
sir please check third line P(h,(c/h)) it should be P(ch,(c/h))
$${sir}\:{please}\:{check}\:{third}\:{line} \\ $$$${P}\left({h},\frac{{c}}{{h}}\right)\:{it}\:{should}\:\:{be}\:{P}\left({ch},\frac{{c}}{{h}}\right) \\ $$
Commented by mr W last updated on 23/Dec/18
what is your curve? xy=c or xy=c^2 ? if xy=c as the question says, then the point is P(h,(c/h)) because h×(c/h)=c.
$${what}\:{is}\:{your}\:{curve}? \\ $$$${xy}={c}\:{or}\:{xy}={c}^{\mathrm{2}} \:? \\ $$$$ \\ $$$${if}\:{xy}={c}\:{as}\:{the}\:{question}\:{says},\:{then} \\ $$$${the}\:{point}\:{is}\:{P}\left({h},\frac{{c}}{{h}}\right)\:{because}\:{h}×\frac{{c}}{{h}}={c}. \\ $$
Commented by peter frank last updated on 23/Dec/18
very sorry sir found my mistake.my xy=c^2 insteady of xy=c
$${very}\:{sorry}\:{sir}\:\:{found}\:{my}\:{mistake}.{my} \\ $$$${xy}={c}^{\mathrm{2}} \:{insteady}\:{of}\:{xy}={c} \\ $$
Answered by peter frank last updated on 23/Dec/18
b) (x^2 /4)−(y^2 /5)=1 a=2 b=(√5) y=±(b/a)x y=±((√5)/2) ii) (((x−1)^2 )/(16))−(((y−3)^2 )/9)=1 a=4 b=3 p=1 q=3 y−q=±(b/a)(x−p) y−3=±(3/4)(x−1) y=(3/4)x−(9/4) or −(3/4)x+(9/4)
$$\left.{b}\right)\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{{y}^{\mathrm{2}} }{\mathrm{5}}=\mathrm{1} \\ $$$${a}=\mathrm{2}\:\:\:{b}=\sqrt{\mathrm{5}} \\ $$$${y}=\pm\frac{{b}}{{a}}{x} \\ $$$${y}=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left.{ii}\right)\:\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{16}}−\frac{\left({y}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1} \\ $$$${a}=\mathrm{4}\:\:\:\:{b}=\mathrm{3}\:\:{p}=\mathrm{1}\:\:\:{q}=\mathrm{3} \\ $$$${y}−{q}=\pm\frac{{b}}{{a}}\left({x}−{p}\right) \\ $$$${y}−\mathrm{3}=\pm\frac{\mathrm{3}}{\mathrm{4}}\left({x}−\mathrm{1}\right) \\ $$$${y}=\frac{\mathrm{3}}{\mathrm{4}}{x}−\frac{\mathrm{9}}{\mathrm{4}}\:{or}\:−\frac{\mathrm{3}}{\mathrm{4}}{x}+\frac{\mathrm{9}}{\mathrm{4}} \\ $$
Answered by peter frank last updated on 23/Dec/18
a) equation of normal at xy=c y−((xt^2 )/c)+(t^3 /c)−(c/t)=0 A(x,0)⇒A(t−(c^2 /t^3 ),0) from tangent at xy=c yt^2 +cx−2ct=0 B(0,y)⇒(0,((2c)/t)) A(t−(c^2 /t^3 ),0),B(0,((2c)/(t ))) x,y)=(((x_1 +x_2 )/2),((y_1 +y_2 )/2)) (x,y)=(((ct)/2)−(c/(2t^3 ))),(c/y) x=(c/2)−(c/(2t^3 ))......(i) t=(c/y).......(ii) sub ii in i 2x=(c/y)−(y^3 /c) y^4 +2xcy−c^2 =0
$$\left.{a}\right)\:{equation}\:{of}\:{normal} \\ $$$${at}\:{xy}={c} \\ $$$${y}−\frac{{xt}^{\mathrm{2}} }{{c}}+\frac{{t}^{\mathrm{3}} }{{c}}−\frac{{c}}{{t}}=\mathrm{0} \\ $$$${A}\left({x},\mathrm{0}\right)\Rightarrow{A}\left({t}−\frac{{c}^{\mathrm{2}} }{{t}^{\mathrm{3}} },\mathrm{0}\right) \\ $$$${from}\:{tangent}\:{at}\:{xy}={c} \\ $$$${yt}^{\mathrm{2}} +{cx}−\mathrm{2}{ct}=\mathrm{0} \\ $$$${B}\left(\mathrm{0},{y}\right)\Rightarrow\left(\mathrm{0},\frac{\mathrm{2}{c}}{{t}}\right) \\ $$$${A}\left({t}−\frac{{c}^{\mathrm{2}} }{{t}^{\mathrm{3}} },\mathrm{0}\right),{B}\left(\mathrm{0},\frac{\mathrm{2}{c}}{{t}\:\:}\right) \\ $$$$\left.{x},{y}\right)=\left(\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} }{\mathrm{2}},\frac{{y}_{\mathrm{1}} +{y}_{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\left({x},{y}\right)=\left(\frac{{ct}}{\mathrm{2}}−\frac{{c}}{\mathrm{2}{t}^{\mathrm{3}} }\right),\frac{{c}}{{y}} \\ $$$${x}=\frac{{c}}{\mathrm{2}}−\frac{{c}}{\mathrm{2}{t}^{\mathrm{3}} }……\left({i}\right) \\ $$$${t}=\frac{{c}}{{y}}…….\left({ii}\right) \\ $$$${sub}\:{ii}\:\:{in}\:{i} \\ $$$$\mathrm{2}{x}=\frac{{c}}{{y}}−\frac{{y}^{\mathrm{3}} }{{c}} \\ $$$${y}^{\mathrm{4}} +\mathrm{2}{xcy}−{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$
Commented by mr W last updated on 23/Dec/18
please check eqn. of normal: y−xt^2 +ct^3 −(c/t)=0 it should be, i think, y−((xt^2 )/c)+(t^3 /c)−(c/t)=0
$${please}\:{check}\:{eqn}.\:{of}\:{normal}: \\ $$$${y}−{xt}^{\mathrm{2}} +{ct}^{\mathrm{3}} −\frac{{c}}{{t}}=\mathrm{0} \\ $$$${it}\:{should}\:{be},\:{i}\:{think}, \\ $$$${y}−\frac{{xt}^{\mathrm{2}} }{{c}}+\frac{{t}^{\mathrm{3}} }{{c}}−\frac{{c}}{{t}}=\mathrm{0} \\ $$
Commented by peter frank last updated on 23/Dec/18
Mrw please check right or wrong
$$\:{Mrw}\:{please}\:{check}\:{right}\:{or}\:{wrong} \\ $$
Commented by mr W last updated on 23/Dec/18
please check eqn. of tangent: yt^2 +x−2ct=0pp it should be, i think, yt^2 +cx−2ct=0
$${please}\:{check}\:{eqn}.\:{of}\:{tangent}: \\ $$$${yt}^{\mathrm{2}} +{x}−\mathrm{2}{ct}=\mathrm{0}{pp} \\ $$$${it}\:{should}\:{be},\:{i}\:{think}, \\ $$$${yt}^{\mathrm{2}} +{cx}−\mathrm{2}{ct}=\mathrm{0} \\ $$
Commented by mr W last updated on 23/Dec/18
please check: the final eqn. should be, i think, y^4 +2cxy−c^2 =0
$${please}\:{check}: \\ $$$${the}\:{final}\:{eqn}.\:{should}\:{be},\:{i}\:{think}, \\ $$$${y}^{\mathrm{4}} +\mathrm{2}{cxy}−{c}^{\mathrm{2}} =\mathrm{0} \\ $$
Commented by peter frank last updated on 23/Dec/18
equation of tangent and normql to xy=c at (ct,(c/t)) x=ct y=(c/t) y^′ =−(1/t^2 ) ((y−(c/t))/(x−ct))=−(1/t^2 ) yt^2 +x−2ct=0 normal to xy=c y^′ =t^2 ((y−(c/t))/(x−ct))=t^2 xt^2 −ct^3 =y−(c/t) xt^2 −ct^3 +(c/t)=y
$${equation}\:{of}\:{tangent}\:{and}\:\:{normql} \\ $$$${to}\:{xy}={c}\:{at}\:\left({ct},\frac{{c}}{{t}}\right) \\ $$$${x}={ct} \\ $$$${y}=\frac{{c}}{{t}} \\ $$$${y}^{'} =−\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$$\frac{{y}−\frac{{c}}{{t}}}{{x}−{ct}}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${yt}^{\mathrm{2}} +{x}−\mathrm{2}{ct}=\mathrm{0} \\ $$$${normal}\:{to}\:{xy}={c} \\ $$$${y}^{'} ={t}^{\mathrm{2}} \\ $$$$\frac{{y}−\frac{{c}}{{t}}}{{x}−{ct}}={t}^{\mathrm{2}} \\ $$$${xt}^{\mathrm{2}} −{ct}^{\mathrm{3}} ={y}−\frac{{c}}{{t}} \\ $$$${xt}^{\mathrm{2}} −{ct}^{\mathrm{3}} +\frac{{c}}{{t}}={y} \\ $$$$ \\ $$
Commented by peter frank last updated on 23/Dec/18
Mrw sir please check
$${Mrw}\:{sir}\:{please}\:{check} \\ $$
Commented by mr W last updated on 23/Dec/18
if xy=c, then at point (t,(c/t)), not at (ct,(c/t)) y′=−(c/x^2 )=−(c/t^2 ) ......
$${if}\:{xy}={c},\:{then} \\ $$$${at}\:{point}\:\left({t},\frac{{c}}{{t}}\right),\:{not}\:{at}\:\left({ct},\frac{{c}}{{t}}\right) \\ $$$${y}'=−\frac{{c}}{{x}^{\mathrm{2}} }=−\frac{{c}}{{t}^{\mathrm{2}} } \\ $$$$…… \\ $$
Commented by peter frank last updated on 23/Dec/18
your absolute right
$${your}\:{absolute}\:{right} \\ $$

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