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Question Number 165431 by ajfour last updated on 01/Feb/22

	Answered by mr W last updated on 01/Feb/22

	Commented by mr W last updated on 01/Feb/22

	$P (p, a^{2} - p^{2})$ $Q (h, r)$ $\tan θ = 2 p$ $\Rightarrow \cos θ = \frac{1}{\sqrt{1 + 4 p^{2}}}, \sin θ = \frac{2 p}{\sqrt{1 + 4 p^{2}}}$ $r = a^{2} - p^{2} + r \cos θ$ $\Rightarrow r = \frac{a^{2} - p^{2}}{1 - \frac{1}{\sqrt{1 + 4 p^{2}}}}$ $h = p + r \sin θ = p + \frac{2 p r}{\sqrt{1 + 4 p^{2}}}$ $t a n g e n t l i n e : y = c + m x$ $m = \frac{1}{\tan θ} = \frac{1}{2 p}$ $y = c + \frac{x}{2 p}$ $c + \frac{x}{2 p} = a^{2} - x^{2}$ $x^{2} + \frac{x}{2 p} + c - a^{2} = 0$ $Δ = \frac{1}{4 p^{2}} - 4 (c - a^{2}) = 0$ $\Rightarrow c = \frac{1}{16 p^{2}} + a^{2}$ $\frac{x}{2 p} - y + \frac{1}{16 p^{2}} + a^{2} = 0$ $r = \frac{∣ \frac{1}{2 p} (p + \frac{2 p r}{\sqrt{1 + 4 p^{2}}}) - r + \frac{1}{16 p^{2}} + a^{2} ∣}{\sqrt{1 + \frac{1}{4 p^{2}}}}$ $r \sqrt{1 + \frac{1}{4 p^{2}}} = \frac{1}{2 p} (p + \frac{2 p r}{\sqrt{1 + 4 p^{2}}}) - r + \frac{1}{16 p^{2}} + a^{2}$ $\Rightarrow \frac{(a^{2} - p^{2}) \sqrt{1 + 4 p^{2}}}{\sqrt{1 + 4 p^{2}} - 1} (\frac{\sqrt{1 + 4 p^{2}}}{2 p} - \frac{1}{\sqrt{1 + 4 p^{2}}} + 1) = \frac{1}{2} + \frac{1}{16 p^{2}} + a^{2}$
	Commented by mr W last updated on 01/Feb/22

	Commented by ajfour last updated on 01/Feb/22

	$B u t s i r, i n y o u r s o l u t i o n$ $r = f (p), p i s n t d e t e r m i n e d$ $t h o u g h . . .$
	Commented by mr W last updated on 01/Feb/22

	$i c a n o n l y g e t a n e q u a t i o n f o r p,$ $w h i c h c a n b e s o l v e d n u m e r i c a l l y$ $f o r a g i v e n a . w i t h t h i s p w e c a n$ $c a l c u l a t e r .$ $i {c a n}^{'} t f i n d a n e q u a t i o n d i r e c t l y f o r$ $r .$
	Commented by mr W last updated on 01/Feb/22

	$o r w e c a n t a k e p a s p a r a m e t e r a n d$ $s h o w t h e r e l a t i o n s h i p b e t w e e n a$ $a n d r :$
	Commented by mr W last updated on 01/Feb/22

	Commented by ajfour last updated on 01/Feb/22
	Yeah, even i tried n got two eqns. in r, p, a. Thanks sir, i shall chase it bit more..
	Commented by Tawa11 last updated on 02/Feb/22

	$Great sir$
	Answered by ajfour last updated on 02/Feb/22

	Commented by ajfour last updated on 02/Feb/22

	$y = \frac{x}{2 p} + c = a^{2} - x^{2} (\tan θ = \frac{1}{2 p})$ $\Rightarrow x^{2} + \frac{x}{2 p} + c - a^{2} = 0$ $x = p = - \frac{1}{4 p} + \sqrt{\frac{1}{16 p^{2}} + a^{2} - c}$ $\Rightarrow p^{2} + \frac{1}{2} = a^{2} - c$ $\Rightarrow c = a^{2} - (\frac{1}{2} + p^{2}) . . (i)$ $N o w$ $r - r \sin θ = r - \frac{r}{\sqrt{1 + 4 p^{2}}}$ $= a^{2} - p^{2}$ $\Rightarrow r = \frac{a^{2} - p^{2}}{1 - \frac{1}{\sqrt{1 + 4 p^{2}}}} . . . (i i)$ $c o n s i d e r u p p e r l i n e$ $y = \frac{x}{2 p} + c + r \sec θ [n o w w i t h . . (i)]$ $a^{2} - x^{2} =$ $\frac{x}{2 p} + a^{2} - (\frac{1}{2} + p^{2}) + \frac{r \sqrt{1 + 4 p^{2}}}{2 p}$ $D = 0 \Rightarrow$ $\frac{1}{4 p^{2}} = \frac{2 r \sqrt{1 + 4 p^{2}}}{p} - (2 + 4 p^{2})$ $u s i n g . . (i i)$ $\frac{1}{4 p^{2}} = \frac{2 \sqrt{1 + 4 p^{2}}}{p} (\frac{a^{2} - p^{2}}{1 - \frac{1}{\sqrt{1 + 4 p^{2}}}}) - (2 + 4 p^{2})$ $\Rightarrow$ $(2 p + \frac{1}{2 p}) \sqrt{2 p} = \frac{4 (a^{2} - p^{2})}{\sqrt{\frac{1}{2 p} + 2 p} - \frac{1}{\sqrt{2 p}}}$ $A n d r = \frac{a^{2} - p^{2}}{1 - \frac{1}{\sqrt{1 + 4 p^{2}}}} .$
	Commented by ajfour last updated on 02/Feb/22

	$a = 2, p \approx 1.4906, r \approx 2.60728$ $a = 3, p \approx 2.21163, r \approx 5.27102$
	Commented by mr W last updated on 02/Feb/22

	$v e r y f i n e s i r!$
	Commented by ajfour last updated on 02/Feb/22

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