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Question Number 80206 by peter frank last updated on 01/Feb/20
Commented by mr W last updated on 01/Feb/20
the language of the question is not to understand. please make the question clear! what exact is to prove?
$${the}\:{language}\:{of}\:{the}\:{question}\:{is}\:{not} \\ $$$${to}\:{understand}. \\ $$$${please}\:{make}\:{the}\:{question}\:{clear}!\:{what} \\ $$$${exact}\:{is}\:{to}\:{prove}?\: \\ $$
Commented by mr W last updated on 01/Feb/20
Answered by mr W last updated on 01/Feb/20
say the tangent passing the given point is y−(√(a^2 +b^2 ))=m(x−(a^2 /( (√(a^2 −b^2 ))))) y=mx+(√(a^2 +b^2 ))−((ma^2 )/( (√(a^2 −b^2 )))) we have m^2 a^2 +b^2 =((√(a^2 +b^2 ))−((ma^2 )/( (√(a^2 −b^2 )))))^2 m^2 a^2 +b^2 =a^2 +b^2 +((m^2 a^4 )/(a^2 −b^2 ))−((2ma^2 (√(a^2 +b^2 )))/( (√(a^2 −b^2 )))) ((m^2 b^2 )/(a^2 −b^2 ))−((2m(√(a^2 +b^2 )))/( (√(a^2 −b^2 ))))+1=0 b^2 m^2 −2m(√(a^4 −b^4 ))+a^2 −b^2 =0 ⇒m=((((√(a^2 +b^2 ))±a)(√(a^2 −b^2 )))/b^2 ) tangent line 1: y−(√(a^2 +b^2 ))=[((((√(a^2 +b^2 ))−a)(√(a^2 −b^2 )))/b^2 )](x−(a^2 /( (√(a^2 −b^2 ))))) intersection with y−axis at point P: y_P −(√(a^2 +b^2 ))=[((((√(a^2 +b^2 ))−a)(√(a^2 −b^2 )))/b^2 )](−(a^2 /( (√(a^2 −b^2 ))))) ⇒y_P =((a^2 /b^2 ))a−((a^2 /b^2 )−1)(√(a^2 +b^2 )) tangent line 2: y−(√(a^2 +b^2 ))=[((((√(a^2 +b^2 ))+a)(√(a^2 −b^2 )))/b^2 )](x−(a^2 /( (√(a^2 −b^2 ))))) intersection with x−axis at point Q: 0−(√(a^2 +b^2 ))=[((((√(a^2 +b^2 ))+a)(√(a^2 −b^2 )))/b^2 )](x_Q −(a^2 /( (√(a^2 −b^2 ))))) ⇒x_Q =((a(√(a^2 +b^2 ))−b^2 )/( (√(a^2 −b^2 )))) ......
$${say}\:{the}\:{tangent}\:{passing}\:{the}\:{given}\:{point}\:{is} \\ $$$${y}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={m}\left({x}−\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right) \\ $$$${y}={mx}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\frac{{ma}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$$${we}\:{have} \\ $$$${m}^{\mathrm{2}} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\frac{{ma}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right)^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\frac{{m}^{\mathrm{2}} {a}^{\mathrm{4}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }−\frac{\mathrm{2}{ma}^{\mathrm{2}} \sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$$$\frac{{m}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }−\frac{\mathrm{2}{m}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}+\mathrm{1}=\mathrm{0} \\ $$$${b}^{\mathrm{2}} {m}^{\mathrm{2}} −\mathrm{2}{m}\sqrt{{a}^{\mathrm{4}} −{b}^{\mathrm{4}} }+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{m}=\frac{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\pm{a}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} } \\ $$$${tangent}\:{line}\:\mathrm{1}: \\ $$$${y}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\left[\frac{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−{a}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} }\right]\left({x}−\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right) \\ $$$${intersection}\:{with}\:{y}−{axis}\:{at}\:{point}\:{P}: \\ $$$${y}_{{P}} −\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\left[\frac{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−{a}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} }\right]\left(−\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{y}_{{P}} =\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right){a}−\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right)\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${tangent}\:{line}\:\mathrm{2}: \\ $$$${y}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\left[\frac{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+{a}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} }\right]\left({x}−\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right) \\ $$$${intersection}\:{with}\:{x}−{axis}\:{at}\:{point}\:{Q}: \\ $$$$\mathrm{0}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\left[\frac{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+{a}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} }\right]\left({x}_{{Q}} −\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{x}_{{Q}} =\frac{{a}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−{b}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$$$…… \\ $$
Commented by peter frank last updated on 01/Feb/20
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$
Commented by mr W last updated on 01/Feb/20
solution is not complete yet! it can not be completed since it′s not clear what should be proved. where did you get the question? can you check the question once more?
$${solution}\:{is}\:{not}\:{complete}\:{yet}!\:{it}\:{can} \\ $$$${not}\:{be}\:{completed}\:{since}\:{it}'{s}\:{not}\:{clear} \\ $$$${what}\:{should}\:{be}\:{proved}.\: \\ $$$${where}\:{did}\:{you}\:{get}\:{the}\:{question}?\:{can} \\ $$$${you}\:{check}\:{the}\:{question}\:{once}\:{more}? \\ $$
Commented by peter frank last updated on 01/Feb/20
I check the question on the paper it seems no problem on typing may language used
$${I}\:{check}\:{the}\:{question}\:{on}\:{the}\:{paper}\:{it}\:{seems}\:{no}\:{problem}\:{on}\:{typing}\:\:{may} \\ $$$${language}\:{used} \\ $$$$ \\ $$
Commented by mr W last updated on 01/Feb/20
Commented by mr W last updated on 01/Feb/20
if you understand this, please explain me what is meant with it.
$${if}\:{you}\:{understand}\:{this},\:{please}\:{explain} \\ $$$${me}\:{what}\:{is}\:{meant}\:{with}\:{it}. \\ $$
Commented by peter frank last updated on 01/Feb/20
i think there two case inside case 1 interception of ordinate through the focus case 2 intercept on the ordinate near distance equal to major axis
$${i}\:{think}\:{there}\:{two}\:{case}\:{inside} \\ $$$${case}\:\mathrm{1} \\ $$$${interception}\:{of}\:{ordinate} \\ $$$${through}\:{the}\:{focus} \\ $$$${case}\:\mathrm{2} \\ $$$${intercept}\:{on}\:{the}\:{ordinate} \\ $$$${near}\:{distance}\:{equal}\:{to}\: \\ $$$${major}\:{axis} \\ $$
Commented by mr W last updated on 01/Feb/20
i totally don′t understand what you are saying, maybe you either. what should be proved? i have given a diagram above in the comment to the question, the diagram shown the situation. please use this diagram for your explanation. if you also don′t understand the question, just say it, and we forget the question.
$${i}\:{totally}\:{don}'{t}\:{understand}\:{what}\:{you} \\ $$$${are}\:{saying},\:{maybe}\:{you}\:{either}. \\ $$$${what}\:{should}\:{be}\:{proved}? \\ $$$${i}\:{have}\:{given}\:{a}\:{diagram}\:{above} \\ $$$${in}\:{the}\:{comment}\:{to}\:{the}\:{question}, \\ $$$${the}\:{diagram}\:{shown}\:{the}\:{situation}. \\ $$$${please}\:{use}\:{this}\:{diagram}\:{for}\:{your} \\ $$$${explanation}.\:{if}\:{you}\:{also}\:{don}'{t} \\ $$$${understand}\:{the}\:{question},\:{just}\:{say}\:{it}, \\ $$$${and}\:{we}\:{forget}\:{the}\:{question}. \\ $$
Commented by peter frank last updated on 01/Feb/20
to be honest sir i dont understand.that why my explanation was out of point
$${to}\:{be}\:{honest}\:{sir}\:{i}\:{dont} \\ $$$${understand}.{that}\:{why}\:{my} \\ $$$${explanation}\:{was}\:{out}\:{of}\:{point} \\ $$
Commented by mr W last updated on 01/Feb/20
alright, i see.
$${alright},\:{i}\:{see}. \\ $$

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