Question-125800

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

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

\dots

Answered by peter frank last updated on 14/Dec/20