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Question Number 43331 by ajfour last updated on 09/Sep/18

Commented by ajfour last updated on 09/Sep/18

$F i n d e q u a t i o n o f r e d p a r a b o l a,$ $i f i t b e r e f l e c t i o n o f y = x^{2} i n$ $t h e l i n e y = m x + c .$
Commented by ajfour last updated on 09/Sep/18

Commented by ajfour last updated on 09/Sep/18

$p = \frac{c + m h - k}{\sqrt{1 + m^{2}}}$ $y - k = 2 p \cos θ$ $h - x = 2 p \sin θ$ $\tan θ = m$ $\Rightarrow x = h - \frac{2 m (c + m h - k)}{1 + m^{2}}$ $a n d y = k + \frac{2 (c + m h - k)}{1 + m^{2}}$ $L e t s c o n s i d e r a p o i n t P (h, k) o f$ $r e d p a r a b o l a .$ $\Rightarrow f (h, k) = 0$ $l e t i m a g e o f P (h, k) b e I (x, y)$ $B u t y = x^{2}, \Rightarrow$ $k + \frac{2 (c + m h - k)}{1 + m^{2}} = {[h - \frac{2 m (c + m h - k)}{1 + m^{2}}]}^{2}$ $h e n c e f (x, y) i s$ $y + \frac{2 (c + m x - y)}{1 + m^{2}} = {[x - \frac{2 m (c + m x - y)}{1 + m^{2}}]}^{2} .$ $J u s t t o c h e c k$ $l e t m = 1, c = 0$ $\Rightarrow x = y^{2} (r e f l e c t i o n o f y = x^{2}) .$
Commented by ajfour last updated on 10/Sep/18

$T h a n k y o u S i r .$
Commented by MrW3 last updated on 09/Sep/18

$c o r r e c t s i r .$ $G e n e r a l l y m i r r o r i m a g e o f p o i n t P (x, y) i n$ $l i n e a x + b y + c = 0 i s p o i n t Q (u, v) w i t h$ $u = x - \frac{2 a (a x + b y + c)}{a^{2} + b^{2}}$ $v = y - \frac{2 b (a x + b y + c)}{a^{2} + b^{2}}$
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