Math Question and Answers


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 119314 by peter frank last updated on 23/Oct/20

Commented by mr W last updated on 23/Oct/20

$w h a t i s a ?$
Commented by mr W last updated on 23/Oct/20

$o r d o y o u m e a n 3 y + a = 0 ?$
Commented by peter frank last updated on 23/Oct/20

$costant$
Commented by mr W last updated on 23/Oct/20

$i k n o w i t i s a c o n s t a n t, b u t$ $w h a t d o e s i t m e a n ?$
Commented by peter frank last updated on 23/Oct/20

$I dont know sir$
	Answered by mr W last updated on 23/Oct/20

	Commented by mr W last updated on 23/Oct/20

	$s a y v e r t e x i s t h e o r i g i n, t h e f a c u s i s$ $F (0, h), t h e n t h e e q n . o f p a r a b o l a i s$ $y = \frac{x^{2}}{4 h}$ $s a y t h e f o c a l c h o r d i s$ $y = h + m x w i t h m = \tan θ$ $i n t e r s e c t i o n w i t h p a r a b o l a :$ $\frac{x^{2}}{4 h} = h + m x$ $x^{2} - 4 m h x - 4 h^{2} = 0$ $\Rightarrow x = 2 h (m \pm \sqrt{m^{2} + 1})$ $e n d p o i n t A :$ $x_{A} = 2 h (m - \sqrt{m^{2} + 1})$ $y_{A} = h + 2 h m (m - \sqrt{m^{2} + 1})$ $t a n g e n t a t A :$ $\frac{d y}{d x} = \frac{x}{2 h} = m - \sqrt{m^{2} + 1}$ $y = h + 2 h m (m - \sqrt{m^{2} + 1}) + (m - \sqrt{m^{2} + 1}) [x - 2 h (m - \sqrt{m^{2} + 1})]$ $0 = h + 2 h m (m - \sqrt{m^{2} + 1}) + (m - \sqrt{m^{2} + 1}) [x - 2 h (m - \sqrt{m^{2} + 1})]$ $\Rightarrow x = x_{Q} = - \frac{h}{m - \sqrt{m^{2} + 1}} - 2 h \sqrt{m^{2} + 1}$ $e n d p o i n t B :$ $x_{B} = 2 h (m + \sqrt{m^{2} + 1})$ $y_{B} = h + 2 h m (m + \sqrt{m^{2} + 1})$ $t a n g e n t a t B :$ $y = h + 2 h m (m + \sqrt{m^{2} + 1}) + (m + \sqrt{m^{2} + 1}) [x - 2 h (m + \sqrt{m^{2} + 1})]$ $0 = h + 2 h m (m + \sqrt{m^{2} + 1}) + (m + \sqrt{m^{2} + 1}) [x - 2 h (m + \sqrt{m^{2} + 1})]$ $\Rightarrow x = x_{R} = - \frac{h}{m + \sqrt{m^{2} + 1}} + 2 h \sqrt{m^{2} + 1}$ $i n t e r s e c t i o n p o i n t P :$ $y = h + 2 h m (m - \sqrt{m^{2} + 1}) + (m - \sqrt{m^{2} + 1}) [x - 2 h (m - \sqrt{m^{2} + 1})]$ $y = h + 2 h m (m + \sqrt{m^{2} + 1}) + (m + \sqrt{m^{2} + 1}) [x - 2 h (m + \sqrt{m^{2} + 1})]$ $h [{(m + \sqrt{m^{2} + 1})}^{2} - {(m - \sqrt{m^{2} + 1})}^{2}] = (2 h m + x) \sqrt{m^{2} + 1}$ $\Rightarrow x_{P} = 2 m h$ $\Rightarrow y_{P} = - h$ $c e n t r o i d G :$ $y_{G} = \frac{1}{3} y_{P} = - \frac{h}{3} = c o n s t a n t$ $\Rightarrow 3 y_{G} + h = 0 o r 3 y + h = 0$ $x_{G} = \frac{1}{3} (x_{P} + x_{Q} + x_{R})$ $= \frac{1}{3} (2 m h - \frac{h}{m - \sqrt{m^{2} + 1}} - 2 h \sqrt{m^{2} + 1} - \frac{h}{m + \sqrt{m^{2} + 1}} + 2 h \sqrt{m^{2} + 1})$ $= \frac{h}{3} (2 m + \frac{1}{\sqrt{m^{2} + 1} - m} - \frac{1}{\sqrt{m^{2} + 1} + m})$ $= \frac{4 m h}{3} \neq c o n s t a n t$
	Commented by peter frank last updated on 24/Oct/20

	$thank you . understood every line$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum