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Question-125800

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Question Number 125800 by TITA last updated on 13/Dec/20
Commented by TITA last updated on 13/Dec/20
prove
$$\mathrm{prove} \\ $$
Commented by liberty last updated on 14/Dec/20
b^2 x^2 −a^2 (mx+c)^2 =a^2 b^2 b^2 x^2 −a^2 (m^2 x^2 +2mcx+c^2 )−a^2 b^2 =0 (b^2 −a^2 m^2 )x^2 −2a^2 mcx−(a^2 c^2 +a^2 b^2 )=0 D=0 ⇒ 4a^4 m^2 c^2 +4(b^2 −a^2 m^2 )(a^2 c^2 +a^2 b^2 )=0 ⇒a^4 m^2 c^2 +b^2 a^2 c^2 +a^2 b^4 −a^4 m^2 c^2 −a^4 m^2 b^2 =0 ⇒b^2 a^2 c^2 +a^2 b^4 −a^4 m^2 b^2 =0 ⇒a^2 b^2 (c^2 +b^2 −a^2 m^2 )=0 ; a^2 ≠ 0 ∧b^2 ≠0 ⇒c^2 =a^2 m^2 −b^2 ∧ c = (√(a^2 m^2 −b^2 )) .
$${b}^{\mathrm{2}} {x}^{\mathrm{2}} −{a}^{\mathrm{2}} \left({mx}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} −{a}^{\mathrm{2}} \left({m}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{mcx}+{c}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} {m}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {mcx}−\left({a}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${D}=\mathrm{0}\:\Rightarrow\:\mathrm{4}{a}^{\mathrm{4}} {m}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{4}\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} {m}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{4}} {m}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {a}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{4}} −{a}^{\mathrm{4}} {m}^{\mathrm{2}} {c}^{\mathrm{2}} −{a}^{\mathrm{4}} {m}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} {a}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{4}} −{a}^{\mathrm{4}} {m}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} {m}^{\mathrm{2}} \right)=\mathrm{0}\:;\:{a}^{\mathrm{2}} \neq\:\mathrm{0}\:\wedge{b}^{\mathrm{2}} \neq\mathrm{0} \\ $$$$\Rightarrow{c}^{\mathrm{2}} \:={a}^{\mathrm{2}} {m}^{\mathrm{2}} −{b}^{\mathrm{2}} \:\wedge\:{c}\:=\:\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:.\: \\ $$
Commented by liberty last updated on 14/Dec/20
other way ⇒ gradient tangent line ((2x)/a^2 )−((2yy′)/b^2 ) = 0 ; b^2 x=a^2 y.y′ ⇒y′ = ((b^2 x)/(a^2 y)) = m ⇒x=((ma^2 y)/b^2 ) substitute in to hyperbola (((m^2 a^4 y^2 )/b^4 )/a^2 ) − (y^2 /b^2 ) = 1 ; (((m^2 a^2 −b^2 )y^2 )/b^4 )=1 y^2 =(b^4 /(m^2 a^2 −b^2 )) → { ((y=(b^2 /( (√(m^2 a^2 −b^2 )))))),((y=−(b^2 /( (√(m^2 a^2 −b^2 )))))) :} contact point ⇒x=((ma^2 (−(b^2 /( (√(m^2 a^2 −c^2 ))))))/b^2 )=−((ma^2 )/( (√(m^2 a^2 −b^2 )))) thus −(b^2 /( (√(m^2 a^2 −b^2 )))) = ((−m^2 a^2 )/( (√(m^2 a^2 −b^2 )))) + c c = ((m^2 a^2 −b^2 )/( (√(m^2 a^2 −b^2 )))) = (√(m^2 a^2 −b^2 ))
$${other}\:{way}\:\Rightarrow\:{gradient}\:{tangent}\:{line}\: \\ $$$$\frac{\mathrm{2}{x}}{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{yy}'}{{b}^{\mathrm{2}} }\:=\:\mathrm{0}\:;\:{b}^{\mathrm{2}} {x}={a}^{\mathrm{2}} {y}.{y}' \\ $$$$\Rightarrow{y}'\:=\:\frac{{b}^{\mathrm{2}} {x}}{{a}^{\mathrm{2}} {y}}\:=\:{m}\:\Rightarrow{x}=\frac{{ma}^{\mathrm{2}} {y}}{{b}^{\mathrm{2}} } \\ $$$${substitute}\:{in}\:{to}\:{hyperbola} \\ $$$$\frac{\frac{{m}^{\mathrm{2}} {a}^{\mathrm{4}} {y}^{\mathrm{2}} }{{b}^{\mathrm{4}} }}{{a}^{\mathrm{2}} }\:−\:\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1}\:;\:\frac{\left({m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){y}^{\mathrm{2}} }{{b}^{\mathrm{4}} }=\mathrm{1} \\ $$$${y}^{\mathrm{2}} =\frac{{b}^{\mathrm{4}} }{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\rightarrow\begin{cases}{{y}=\frac{{b}^{\mathrm{2}} }{\:\sqrt{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}}\\{{y}=−\frac{{b}^{\mathrm{2}} }{\:\sqrt{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}}\end{cases} \\ $$$${contact}\:{point}\:\Rightarrow{x}=\frac{{ma}^{\mathrm{2}} \left(−\frac{{b}^{\mathrm{2}} }{\:\sqrt{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{c}^{\mathrm{2}} }}\right)}{{b}^{\mathrm{2}} }=−\frac{{ma}^{\mathrm{2}} }{\:\sqrt{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$$${thus}\:−\frac{{b}^{\mathrm{2}} }{\:\sqrt{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\:=\:\frac{−{m}^{\mathrm{2}} {a}^{\mathrm{2}} }{\:\sqrt{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\:+\:{c}\: \\ $$$$\:{c}\:=\:\frac{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\:\sqrt{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\:=\:\sqrt{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\: \\ $$$$ \\ $$
Commented by TITA last updated on 14/Dec/20
thamks guys
$$\mathrm{thamks}\:\mathrm{guys} \\ $$
Commented by TITA last updated on 14/Dec/20
thanks
$$\mathrm{thanks} \\ $$
Answered by TITA last updated on 13/Dec/20
please help
$$\mathrm{please}\:\mathrm{help} \\ $$
Answered by peter frank last updated on 14/Dec/20
y=mx+c (x^2 /a^2 )−(y^2 /b^2 )=1 (x^2 /a^2 )−(((mx+c)^2 )/b^2 )=1 put discrimint =0 ...
$$\mathrm{y}=\mathrm{mx}+\mathrm{c} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }−\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }−\frac{\left(\mathrm{mx}+\mathrm{c}\right)^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\mathrm{put}\:\:\mathrm{discrimint}\:=\mathrm{0} \\ $$$$… \\ $$
Answered by peter frank last updated on 14/Dec/20

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