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Question Number 46813 by peter frank last updated on 31/Oct/18

Commented by peter frank last updated on 02/Nov/18

$help please$
	Answered by ajfour last updated on 04/Nov/18

	$e q . o f a s y m p t o t e s :$ $\tan θ = \frac{y}{x} = \pm \frac{b}{a}$ $l e t e q . o f l i n e P Q R :$ $y = m (x - h) + k$ $h = a \sec ϕ, k = b \tan ϕ$ $m = \tan α$ $f o r x_{Q} :$ $\frac{{b x}_{Q}}{a} = m (x_{Q} - h) + k$ $\Rightarrow x_{Q} = \frac{k - m h}{\frac{b}{a} - m}$ $r_{1} = (h - x_{Q}) \sec α$ $= (\frac{\frac{b h}{a} - k}{\frac{b}{a} - m}) \sec α$ $s i m i l a r l y$ $r_{2} = (h - x_{R}) \sec α$ $= (\frac{\frac{b h}{a} + k}{\frac{b}{a} + m}) \sec α$ $r_{1} r_{2} = [\frac{{(\frac{b h}{a})}^{2} - k^{2}}{{(\frac{b}{a})}^{2} - m^{2}}] \sec^{2} α$ $= [\frac{\frac{h^{2}}{a^{2}} - \frac{k^{2}}{b^{2}}}{\frac{1}{a^{2}} - \frac{m^{2}}{b^{2}}}] (1 + m^{2})$ $b u t P (h, k) l i e s o n h y p e r b o l a$ $s o \frac{h^{2}}{a^{2}} - \frac{k^{2}}{b^{2}} = 1, h e n c e$ $\Rightarrow r_{1} r_{2} = \frac{a^{2} b^{2} (1 + m^{2})}{b^{2} - a^{2} m^{2}} .$
	Commented by peter frank last updated on 04/Nov/18

	$thank you sir$
	Answered by MrW3 last updated on 04/Nov/18

	$l e t p o i n t P b e (h, k)$ $\frac{h^{2}}{a^{2}} - \frac{k^{2}}{b^{2}} = 1$ $l i n e t h r o u g h p o i n t P w i t h s l o p e m :$ $y - k = m (x - h)$ $e q n . o f a s y m p t o t e 1 :$ $y = - \frac{b}{a} x$ $y_{q} = - \frac{b}{a} x_{q}$ $y_{q} - k = m (x_{q} - h)$ $\Rightarrow k = - \frac{b}{a} x_{q} - m (x_{q} - h)$ $\Rightarrow (\frac{b}{a} + m) x_{q} = m h - k$ $\Rightarrow x_{q} = \frac{a (m h - k)}{m a + b}$ $\Rightarrow y_{q} = - \frac{b (m h - k)}{m a + b}$ $e q n . o f a s y m p t o t e 2 :$ $y = \frac{b}{a} x$ $y_{r} = \frac{b}{a} x_{r}$ $y_{r} - k = m (x_{r} - h)$ $\Rightarrow k = \frac{b}{a} x_{r} - m (x_{r} - h)$ $\Rightarrow (\frac{b}{a} - m) x_{r} = - (m h - k)$ $\Rightarrow x_{r} = \frac{a (m h - k)}{a m - b}$ $\Rightarrow y_{r} = \frac{b (m h - k)}{a m - b}$ $Q P = \sqrt{{(x_{q} - h)}^{2} + {(y_{q} - k)}^{2}} = \sqrt{{(\frac{a (m h - k)}{m a + b} - h)}^{2} + {(- \frac{b (m h - k)}{m a + b} - k)}^{2}}$ $\Rightarrow Q P =∣ \frac{(b h + a k) \sqrt{(1 + m^{2})}}{m a + b} ∣$ $R P = \sqrt{{(x_{r} - h)}^{2} + {(y_{r} - k)}^{2}} = \sqrt{{(\frac{a (m h - k)}{a m - b} - h)}^{2} + {(\frac{b (m h - k)}{a m - b} - k)}^{2}}$ $\Rightarrow R P =∣ \frac{(b h - a k) \sqrt{(1 + m^{2})}}{m a - b} ∣$ $Q P \times R P =∣ \frac{(b h + a k) \sqrt{(1 + m^{2})}}{m a + b} \times \frac{(b h - a k) \sqrt{(1 + m^{2})}}{m a - b} ∣$ $=∣ \frac{(b^{2} h^{2} - a^{2} k^{2}) (1 + m^{2})}{m^{2} a^{2} - b^{2}} ∣$ $=∣ \frac{(\frac{h^{2}}{a^{2}} - \frac{k^{2}}{b^{2}}) a^{2} b^{2} (1 + m^{2})}{m^{2} a^{2} - b^{2}} ∣$ $= \frac{a^{2} b^{2} (1 + m^{2})}{∣ m^{2} a^{2} - b^{2} ∣}$
	Commented by peter frank last updated on 04/Nov/18

	$mrW 3 . thank you$
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