Question Number 80206 by peter frank last updated on 01/Feb/20
Commented by mr W last updated on 01/Feb/20
$${the}\:{language}\:{of}\:{the}\:{question}\:{is}\:{not} \\ $$$${to}\:{understand}. \\ $$$${please}\:{make}\:{the}\:{question}\:{clear}!\:{what} \\ $$$${exact}\:{is}\:{to}\:{prove}?\: \\ $$
Commented by mr W last updated on 01/Feb/20
Answered by mr W last updated on 01/Feb/20
))) 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 ))) ......](Q80225.png)
$${say}\:{the}\:{tangent}\:{passing}\:{the}\:{given}\:{point}\:{is} \\ $$$${y}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={m}\left({x}−\frac{{a}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right) \\ $$$${y}={mx}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\frac{{ma}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$$${we}\:{have} \\ $$$${m}^{\mathrm{2}} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\frac{{ma}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right)^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\frac{{m}^{\mathrm{2}} {a}^{\mathrm{4}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }−\frac{\mathrm{2}{ma}^{\mathrm{2}} \sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$$$\frac{{m}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }−\frac{\mathrm{2}{m}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}+\mathrm{1}=\mathrm{0} \\ $$$${b}^{\mathrm{2}} {m}^{\mathrm{2}} −\mathrm{2}{m}\sqrt{{a}^{\mathrm{4}} −{b}^{\mathrm{4}} }+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{m}=\frac{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\pm{a}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} } \\ $$$${tangent}\:{line}\:\mathrm{1}: \\ $$$${y}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\left[\frac{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−{a}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} }\right]\left({x}−\frac{{a}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right) \\ $$$${intersection}\:{with}\:{y}−{axis}\:{at}\:{point}\:{P}: \\ $$$${y}_{{P}} −\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\left[\frac{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−{a}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} }\right]\left(−\frac{{a}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{y}_{{P}} =\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right){a}−\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right)\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${tangent}\:{line}\:\mathrm{2}: \\ $$$${y}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\left[\frac{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+{a}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} }\right]\left({x}−\frac{{a}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right) \\ $$$${intersection}\:{with}\:{x}−{axis}\:{at}\:{point}\:{Q}: \\ $$$$\mathrm{0}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\left[\frac{\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+{a}\right)\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}^{\mathrm{2}} }\right]\left({x}_{{Q}} −\frac{{a}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{x}_{{Q}} =\frac{{a}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−{b}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$$$...... \\ $$
Commented by peter frank last updated on 01/Feb/20
$${thank}\:{you}\:{sir}. \\ $$
Commented by mr W last updated on 01/Feb/20
$${solution}\:{is}\:{not}\:{complete}\:{yet}!\:{it}\:{can} \\ $$$${not}\:{be}\:{completed}\:{since}\:{it}'{s}\:{not}\:{clear} \\ $$$${what}\:{should}\:{be}\:{proved}.\: \\ $$$${where}\:{did}\:{you}\:{get}\:{the}\:{question}?\:{can} \\ $$$${you}\:{check}\:{the}\:{question}\:{once}\:{more}? \\ $$
Commented by peter frank last updated on 01/Feb/20
$${I}\:{check}\:{the}\:{question}\:{on}\:{the}\:{paper}\:{it}\:{seems}\:{no}\:{problem}\:{on}\:{typing}\:\:{may} \\ $$$${language}\:{used} \\ $$$$ \\ $$
Commented by mr W last updated on 01/Feb/20
Commented by mr W last updated on 01/Feb/20
$${if}\:{you}\:{understand}\:{this},\:{please}\:{explain} \\ $$$${me}\:{what}\:{is}\:{meant}\:{with}\:{it}. \\ $$
Commented by peter frank last updated on 01/Feb/20
$${i}\:{think}\:{there}\:{two}\:{case}\:{inside} \\ $$$${case}\:\mathrm{1} \\ $$$${interception}\:{of}\:{ordinate} \\ $$$${through}\:{the}\:{focus} \\ $$$${case}\:\mathrm{2} \\ $$$${intercept}\:{on}\:{the}\:{ordinate} \\ $$$${near}\:{distance}\:{equal}\:{to}\: \\ $$$${major}\:{axis} \\ $$
Commented by mr W last updated on 01/Feb/20
$${i}\:{totally}\:{don}'{t}\:{understand}\:{what}\:{you} \\ $$$${are}\:{saying},\:{maybe}\:{you}\:{either}. \\ $$$${what}\:{should}\:{be}\:{proved}? \\ $$$${i}\:{have}\:{given}\:{a}\:{diagram}\:{above} \\ $$$${in}\:{the}\:{comment}\:{to}\:{the}\:{question}, \\ $$$${the}\:{diagram}\:{shown}\:{the}\:{situation}. \\ $$$${please}\:{use}\:{this}\:{diagram}\:{for}\:{your} \\ $$$${explanation}.\:{if}\:{you}\:{also}\:{don}'{t} \\ $$$${understand}\:{the}\:{question},\:{just}\:{say}\:{it}, \\ $$$${and}\:{we}\:{forget}\:{the}\:{question}. \\ $$
Commented by peter frank last updated on 01/Feb/20
$${to}\:{be}\:{honest}\:{sir}\:{i}\:{dont} \\ $$$${understand}.{that}\:{why}\:{my} \\ $$$${explanation}\:{was}\:{out}\:{of}\:{point} \\ $$
Commented by mr W last updated on 01/Feb/20
$${alright},\:{i}\:{see}. \\ $$