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Question Number 164728 by mr W last updated on 21/Jan/22

Commented by mr W last updated on 21/Jan/22

$t w o p o i n t s P a n d Q o n t h e p a r a b o l a$ $y = \frac{x^{2}}{a} w i t h ∣ P Q ∣= l . t h e n o r m a l l i n e s$ $a t P a n d Q i n t e r s e c t a t p o i n t S .$ $f i n d t h e l o c u s o f S .$
	Answered by ajfour last updated on 21/Jan/22

	$S (- h, k)$ $N o r m a l f r o m S$ $y = - (\frac{a}{2 p}) x + c P (p, \frac{p^{2}}{a}) l i e s o n i t$ $\frac{p^{2}}{a} = - \frac{a}{2} + c$ ${(p_{2} - p_{1})}^{2} + \frac{{(p_{2}^{2} - p_{1}^{2})}^{2}}{a^{3}} = L^{2} . . . (I)$ $k = \frac{a h}{2 p} + \frac{p^{2}}{a} + \frac{a}{2}$ $\Rightarrow 2 p^{3} + a (a - 2 k) p + a h = 0$ $w e g e t p_{1}, p_{2}, p_{3} i n t e r m s o f$ $h, k, a .$ $s u b s t i t u t i n g i n . . (I) g i v e s r e q u i r e d$ $t h e t h r e e l o c i i, f o r t h r e e c h o i c e s o f$ $p a i r s f r o m a m o n g p_{1}, p_{2}, p_{3} .$
	Commented by mr W last updated on 21/Jan/22

	$t h a n k s f o r t r y i n g s i r! i^{'} l l a l s o g i v e a$ $t r y .$
	Answered by mr W last updated on 21/Jan/22

	Commented by mr W last updated on 23/Jan/22

	$l o c u s i n s h a p e o f a G e r m a n B r e z e l . . .$
	Commented by mr W last updated on 22/Jan/22

	Commented by mr W last updated on 21/Jan/22
	$say P(p, (p^2 /a)), Q(q, (q^2 /a)) tan φ=((2p)/a) tan ϕ=((2q)/a) (q−p)^2 +((q^2 /a)−(p^2 /a))^2 =l^2 (q−p)^2 [1+(((p+q)^2 )/a^2 )]=l^2 [(q+p)^2 −4pq][1+(((p+q)^2 )/a^2 )]=l^2 let u=p+q, v=pq (u^2 −4v)(1+(u^2 /a^2 ))=l^2 ⇒v=(1/4)(u^2 −((a^2 l^2 )/(a^2 +u^2 ))) eqn. of PS: y=(p^2 /a)−(a/(2p))(x−p) y=(p^2 /a)+(a/2)−((ax)/(2p)) eqn. of QS: y=(q^2 /a)+(a/2)−((ax)/(2q)) say intersection point S(x_s ,y_s ) (p^2 /a)+(a/2)−((ax_s )/(2p))=(q^2 /a)+(a/2)−((ax_s )/(2q)) (p^2 /a)−((ax_s )/(2p))=(q^2 /a)−((ax_s )/(2q)) x_s =−((2(q+p)pq)/a^2 ) ⇒x_s =−((2uv)/a^2 ) 2y_s =((p^2 +q^2 )/a)+a−ax_s ((1/(2p))+(1/(2q))) y_s =(a/2)+((p^2 +q^2 )/(2a))−((a(p+q)x_s )/(4pq)) y_s =(a/2)+(((p+q)^2 −pq)/a) ⇒y_s =(a/2)+((u^2 −v)/a) taking u as parameter we get the eqn. of locus of point S: { ((x=−((2uv)/a^2 ))),((y=(a/2)+((u^2 −v)/a))) :} with v=(1/4)(u^2 −((a^2 l^2 )/(a^2 +u^2 ))) with u given, p, q are roots of z^2 −uz+(1/4)(u^2 −((a^2 l^2 )/(a^2 +u^2 )))=0 ⇒p,q=(1/2)(u∓((al)/( (√(a^2 +u^2 )))))$
	$s a y P (p, \frac{p^{2}}{a}), Q (q, \frac{q^{2}}{a})$ $\tan ϕ = \frac{2 p}{a}$ $\tan φ = \frac{2 q}{a}$ ${(q - p)}^{2} + {(\frac{q^{2}}{a} - \frac{p^{2}}{a})}^{2} = l^{2}$ ${(q - p)}^{2} [1 + \frac{{(p + q)}^{2}}{a^{2}}] = l^{2}$ $[{(q + p)}^{2} - 4 p q] [1 + \frac{{(p + q)}^{2}}{a^{2}}] = l^{2}$ $l e t u = p + q, v = p q$ $(u^{2} - 4 v) (1 + \frac{u^{2}}{a^{2}}) = l^{2}$ $\Rightarrow v = \frac{1}{4} (u^{2} - \frac{a^{2} l^{2}}{a^{2} + u^{2}})$ $e q n . o f P S :$ $y = \frac{p^{2}}{a} - \frac{a}{2 p} (x - p)$ $y = \frac{p^{2}}{a} + \frac{a}{2} - \frac{a x}{2 p}$ $e q n . o f Q S :$ $y = \frac{q^{2}}{a} + \frac{a}{2} - \frac{a x}{2 q}$ $s a y i n t e r s e c t i o n p o i n t S (x_{s}, y_{s})$ $\frac{p^{2}}{a} + \frac{a}{2} - \frac{{a x}_{s}}{2 p} = \frac{q^{2}}{a} + \frac{a}{2} - \frac{{a x}_{s}}{2 q}$ $\frac{p^{2}}{a} - \frac{{a x}_{s}}{2 p} = \frac{q^{2}}{a} - \frac{{a x}_{s}}{2 q}$ $x_{s} = - \frac{2 (q + p) p q}{a^{2}}$ $\Rightarrow x_{s} = - \frac{2 u v}{a^{2}}$ $2 y_{s} = \frac{p^{2} + q^{2}}{a} + a - {a x}_{s} (\frac{1}{2 p} + \frac{1}{2 q})$ $y_{s} = \frac{a}{2} + \frac{p^{2} + q^{2}}{2 a} - \frac{a (p + q) x_{s}}{4 p q}$ $y_{s} = \frac{a}{2} + \frac{{(p + q)}^{2} - p q}{a}$ $\Rightarrow y_{s} = \frac{a}{2} + \frac{u^{2} - v}{a}$ $t a k i n g u a s p a r a m e t e r w e g e t t h e e q n .$ $o f l o c u s o f p o i n t S :$ ${\begin{cases} x = - \frac{2 u v}{a^{2}} \\ y = \frac{a}{2} + \frac{u^{2} - v}{a} \end{cases} w i t h v = \frac{1}{4} (u^{2} - \frac{a^{2} l^{2}}{a^{2} + u^{2}})$ $w i t h u g i v e n, p, q a r e r o o t s o f$ $z^{2} - u z + \frac{1}{4} (u^{2} - \frac{a^{2} l^{2}}{a^{2} + u^{2}}) = 0$ $\Rightarrow p, q = \frac{1}{2} (u \mp \frac{a l}{\sqrt{a^{2} + u^{2}}})$
	Commented by mr W last updated on 21/Jan/22

	Commented by mr W last updated on 21/Jan/22

	Commented by ajfour last updated on 21/Jan/22

	$g r e a t w o r k i n g s s i r, i w i l l t r y n$ $f o l l o w . .$
	Commented by Tawa11 last updated on 21/Jan/22

	$Weldone sir$
	Answered by ajfour last updated on 23/Jan/22
	$S(h,k) ((q−h)/(k−(q^2 /a)))=((2p)/a) ⇒ ak−q^2 =(a^2 /2)(((q−h)/p)) ((p−h)/(k−(p^2 /a)))=((2q)/a) ⇒ ak−p^2 =(a^2 /2)(((p−h)/q)) subtracting ⇒ q^2 −p^2 =(a^2 /2)((p/q)−(q/p))+((a^2 h)/2)((1/p)−(1/q)) ⇒ q+p=((a^2 h)/(2pq))−((a^2 (q+p))/(2pq)) ⇒ q+p=((a^2 h)/(a^2 +2pq)) say q+p=s, pq=m ⇒ s=((a^2 h)/(a^2 +2m)) ...(i) Adding 2ak−(p^2 +q^2 )=(a^2 /2)((p/q)+(q/p))−((a^2 h)/2)((1/p)+(1/q)) ⇒ 2ak−s^2 +2m=(a^2 /2)(((s^2 −2m)/m)−((hs)/m)) .....(ii) And s^2 +((s^2 (s^2 −4m))/a^2 )=L^2 ...(iii) ⇒ using (i) and (iii) 1+((s^2 −4m)/a^2 )=(L^2 /h^2 )(1+(m/a^2 )) ⇒ m((L^2 /(a^2 h^2 ))+(4/a^2 ))=1−(L^2 /h^2 ) ⇒ m=((a^2 (1−(L^2 /h^2 )))/((4+(L^2 /h^2 )))) (1/s)=(1/h)+((2m)/(a^2 h))=(1/h)+(2/h)(((1−(L^2 /h^2 ))/(4+(L^2 /h^2 )))) Now substituting for m, s in (ii) ((2ak)/s^2 )+1+((2m)/s^2 )=(a^2 /2)((1/m)−(2/s^2 )−(h/(ms))) {2ak+2a^2 (((h^2 −L^2 )/(4h^2 +L^2 )))+a^2 }{(1/h)+(2/h)(((h^2 −L^2 )/(4h^2 +L^2 )))}^2 +2 =0 And to make it compact say (h/L)=X , (k/L)=Y ((2a)/L)(Y+((X^2 −1)/(4X^2 +1))+(a/L)){1+2(((X^2 −1)/(4X^2 +1)))}^2 +2X^2 =0 or simply (X/(((1/2)+((X^2 −1)/(4X^2 +1)))))=±(((2a)/L))^(1/2) (√(((1−X^2 )/(4X^2 +1))−Y−(a/L))) say for a=1, L=2 (x/({1+((x^2 −4)/(2(x^2 +1)))}))=±(√(((4−x^2 )/(4(x^2 +1)))−(y/2)−(1/2))) ⇒ y=((4−x^2 )/(2(x^2 +1)))−((16(x^2 +1)^2 )/((3x^2 −2)^2 ))−1$
	$S (h, k)$ $\frac{q - h}{k - \frac{q^{2}}{a}} = \frac{2 p}{a} \Rightarrow a k - q^{2} = \frac{a^{2}}{2} (\frac{q - h}{p})$ $\frac{p - h}{k - \frac{p^{2}}{a}} = \frac{2 q}{a} \Rightarrow a k - p^{2} = \frac{a^{2}}{2} (\frac{p - h}{q})$ $s u b t r a c t i n g$ $\Rightarrow q^{2} - p^{2} = \frac{a^{2}}{2} (\frac{p}{q} - \frac{q}{p}) + \frac{a^{2} h}{2} (\frac{1}{p} - \frac{1}{q})$ $\Rightarrow q + p = \frac{a^{2} h}{2 p q} - \frac{a^{2} (q + p)}{2 p q}$ $\Rightarrow q + p = \frac{a^{2} h}{a^{2} + 2 p q}$ $s a y q + p = s, p q = m$ $\Rightarrow s = \frac{a^{2} h}{a^{2} + 2 m} . . . (i)$ $A d d i n g$ $2 a k - (p^{2} + q^{2}) = \frac{a^{2}}{2} (\frac{p}{q} + \frac{q}{p}) - \frac{a^{2} h}{2} (\frac{1}{p} + \frac{1}{q})$ $\Rightarrow 2 a k - s^{2} + 2 m = \frac{a^{2}}{2} (\frac{s^{2} - 2 m}{m} - \frac{h s}{m})$ $. . . . . (i i)$ $A n d s^{2} + \frac{s^{2} (s^{2} - 4 m)}{a^{2}} = L^{2} . . . (i i i)$ $\Rightarrow u s i n g (i) a n d (i i i)$ $1 + \frac{s^{2} - 4 m}{a^{2}} = \frac{L^{2}}{h^{2}} (1 + \frac{m}{a^{2}})$ $\Rightarrow m (\frac{L^{2}}{a^{2} h^{2}} + \frac{4}{a^{2}}) = 1 - \frac{L^{2}}{h^{2}}$ $\Rightarrow m = \frac{a^{2} (1 - \frac{L^{2}}{h^{2}})}{(4 + \frac{L^{2}}{h^{2}})}$ $\frac{1}{s} = \frac{1}{h} + \frac{2 m}{a^{2} h} = \frac{1}{h} + \frac{2}{h} (\frac{1 - \frac{L^{2}}{h^{2}}}{4 + \frac{L^{2}}{h^{2}}})$ $N o w s u b s t i t u t i n g f o r m, s i n (i i)$ $\frac{2 a k}{s^{2}} + 1 + \frac{2 m}{s^{2}} = \frac{a^{2}}{2} (\frac{1}{m} - \frac{2}{s^{2}} - \frac{h}{m s})$ ${2 a k + 2 a^{2} (\frac{h^{2} - L^{2}}{4 h^{2} + L^{2}}) + a^{2}} {\frac{1}{h} + \frac{2}{h} (\frac{h^{2} - L^{2}}{4 h^{2} + L^{2}})}^{2}$ $+ 2 = 0$ $A n d t o m a k e i t c o m p a c t$ $s a y \frac{h}{L} = X, \frac{k}{L} = Y$ $\frac{2 a}{L} (Y + \frac{X^{2} - 1}{4 X^{2} + 1} + \frac{a}{L}) {1 + 2 (\frac{X^{2} - 1}{4 X^{2} + 1})}^{2}$ $+ 2 X^{2} = 0$ $o r s i m p l y$ $\frac{X}{(\frac{1}{2} + \frac{X^{2} - 1}{4 X^{2} + 1})} = \pm {(\frac{2 a}{L})}^{1 / 2} \sqrt{\frac{1 - X^{2}}{4 X^{2} + 1} - Y - \frac{a}{L}}$ $s a y f o r a = 1, L = 2$ $\frac{x}{{1 + \frac{x^{2} - 4}{2 (x^{2} + 1)}}} = \pm \sqrt{\frac{4 - x^{2}}{4 (x^{2} + 1)} - \frac{y}{2} - \frac{1}{2}}$ $\Rightarrow y = \frac{4 - x^{2}}{2 (x^{2} + 1)} - \frac{16 {(x^{2} + 1)}^{2}}{{(3 x^{2} - 2)}^{2}} - 1$
	Commented by ajfour last updated on 23/Jan/22

	$s o m e l i t t l e e r r o r . . . i s h a l l e d i t$ $t o d a y . .$
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