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Question Number 125800 by TITA last updated on 13/Dec/20

Commented by TITA last updated on 13/Dec/20

$prove$
Commented by liberty last updated on 14/Dec/20

$b^{2} x^{2} - a^{2} {(m x + c)}^{2} = a^{2} b^{2}$ $b^{2} x^{2} - a^{2} (m^{2} x^{2} + 2 m c x + c^{2}) - a^{2} b^{2} = 0$ $(b^{2} - a^{2} m^{2}) x^{2} - 2 a^{2} m c x - (a^{2} c^{2} + a^{2} b^{2}) = 0$ $D = 0 \Rightarrow 4 a^{4} m^{2} c^{2} + 4 (b^{2} - a^{2} m^{2}) (a^{2} c^{2} + a^{2} b^{2}) = 0$ $\Rightarrow a^{4} m^{2} c^{2} + b^{2} a^{2} c^{2} + a^{2} b^{4} - a^{4} m^{2} c^{2} - a^{4} m^{2} b^{2} = 0$ $\Rightarrow b^{2} a^{2} c^{2} + a^{2} b^{4} - a^{4} m^{2} b^{2} = 0$ $\Rightarrow a^{2} b^{2} (c^{2} + b^{2} - a^{2} m^{2}) = 0; a^{2} \neq 0 \land b^{2} \neq 0$ $\Rightarrow c^{2} = a^{2} m^{2} - b^{2} \land c = \sqrt{a^{2} m^{2} - b^{2}} .$
Commented by liberty last updated on 14/Dec/20
$other way ⇒ gradient tangent line ((2x)/a^2 )−((2yy′)/b^2 ) = 0 ; b^2 x=a^2 y.y′ ⇒y′ = ((b^2 x)/(a^2 y)) = m ⇒x=((ma^2 y)/b^2 ) substitute in to hyperbola (((m^2 a^4 y^2 )/b^4 )/a^2 ) − (y^2 /b^2 ) = 1 ; (((m^2 a^2 −b^2 )y^2 )/b^4 )=1 y^2 =(b^4 /(m^2 a^2 −b^2 )) → { ((y=(b^2 /( (√(m^2 a^2 −b^2 )))))),((y=−(b^2 /( (√(m^2 a^2 −b^2 )))))) :} contact point ⇒x=((ma^2 (−(b^2 /( (√(m^2 a^2 −c^2 ))))))/b^2 )=−((ma^2 )/( (√(m^2 a^2 −b^2 )))) thus −(b^2 /( (√(m^2 a^2 −b^2 )))) = ((−m^2 a^2 )/( (√(m^2 a^2 −b^2 )))) + c c = ((m^2 a^2 −b^2 )/( (√(m^2 a^2 −b^2 )))) = (√(m^2 a^2 −b^2 ))$
$o t h e r w a y \Rightarrow g r a d i e n t t a n g e n t l i n e$ $\frac{2 x}{a^{2}} - \frac{2 {y y}^{'}}{b^{2}} = 0; b^{2} x = a^{2} y . y^{'}$ $\Rightarrow y^{'} = \frac{b^{2} x}{a^{2} y} = m \Rightarrow x = \frac{{m a}^{2} y}{b^{2}}$ $s u b s t i t u t e i n t o h y p e r b o l a$ $\frac{\frac{m^{2} a^{4} y^{2}}{b^{4}}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1; \frac{(m^{2} a^{2} - b^{2}) y^{2}}{b^{4}} = 1$ $y^{2} = \frac{b^{4}}{m^{2} a^{2} - b^{2}} \to {\begin{cases} y = \frac{b^{2}}{\sqrt{m^{2} a^{2} - b^{2}}} \\ y = - \frac{b^{2}}{\sqrt{m^{2} a^{2} - b^{2}}} \end{cases}$ $c o n t a c t p o i n t \Rightarrow x = \frac{{m a}^{2} (- \frac{b^{2}}{\sqrt{m^{2} a^{2} - c^{2}}})}{b^{2}} = - \frac{{m a}^{2}}{\sqrt{m^{2} a^{2} - b^{2}}}$ $t h u s - \frac{b^{2}}{\sqrt{m^{2} a^{2} - b^{2}}} = \frac{- m^{2} a^{2}}{\sqrt{m^{2} a^{2} - b^{2}}} + c$ $c = \frac{m^{2} a^{2} - b^{2}}{\sqrt{m^{2} a^{2} - b^{2}}} = \sqrt{m^{2} a^{2} - b^{2}}$
Commented by TITA last updated on 14/Dec/20

$thamks guys$
Commented by TITA last updated on 14/Dec/20

$thanks$
	Answered by TITA last updated on 13/Dec/20

	$please help$
	Answered by peter frank last updated on 14/Dec/20

	$y = mx + c$ $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ $\frac{x^{2}}{a^{2}} - \frac{{(mx + c)}^{2}}{b^{2}} = 1$ $put discrimint = 0$ $. . .$
	Answered by peter frank last updated on 14/Dec/20

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