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Question Number 31639 by Tinkutara last updated on 11/Mar/18

	Answered by MJS last updated on 11/Mar/18

	$P \in p a r : (\begin{matrix} p \\ \frac{p^{2}}{4} - \frac{p}{2} + \frac{5}{4} \end{matrix})$ $y^{'} (p) = \frac{p}{2} - \frac{1}{2} (= k of tangent in P)$ $k of normal n in P : - {(\frac{p}{2} - \frac{1}{2})}^{- 1} =$ $= - \frac{2}{p - 1}$ $y = - \frac{2}{p - 1} x + d$ $in P :$ $\frac{p^{2}}{4} - \frac{p}{2} + \frac{5}{4} = - \frac{2 p}{p - 1} + d$ $d = \frac{p^{3} - 3 p^{2} + 15 p - 5}{4 (p - 1)}$ $n_{P} : y = - \frac{2}{p - 1} x + \frac{p^{3} - 3 p^{2} + 15 p - 5}{4 (p - 1)}$ $now looking for Q \in p a r with$ $k (n_{Q}) = k (t_{P})$ $- \frac{2}{q - 1} = \frac{p}{2} - \frac{1}{2}$ $q = 1 - \frac{4}{p - 1}$ $n_{Q} : y = - \frac{2}{q - 1} x + \frac{q^{3} - 3 q^{2} + 15 q - 5}{4 (q - 1)}$ $\Rightarrow$ $n_{Q} : y = \frac{p - 1}{2} x - \frac{p^{3} - 9 p^{2} + 15 p - 15}{2 {(p - 1)}^{2}}$ $n_{P} \cap n_{Q}$ $x = \frac{p^{2} - 5}{2 (p - 1)}; y = \frac{{(p^{2} - 2 p + 5)}^{2}}{4 {(p - 1)}^{2}}$ $\Rightarrow y = x^{2} - 2 x + 5$ $points of intersection lie on$ $a parabola$
	Commented by Tinkutara last updated on 11/Mar/18

	$B u t a n s w e r i s {(x - 1)}^{2} = y - 2$ $I s i t w r o n g i n b o o k ?$
	Commented by MJS last updated on 11/Mar/18

	$. . . that would be$ $y = x^{2} - 2 x + 3$ $I will check it again$
	Commented by MJS last updated on 11/Mar/18

	$I found no mistake in my calculation,$ $please check it again$
	Commented by Tinkutara last updated on 12/Mar/18

	$I s a w a g a i n b u t a n s w e r g i v e n i s o n l y$ ${(x - 1)}^{2} = y - 2 . H o w t o v e r i f y w h i c h i s$ $c o r r e c t ?$
	Commented by MJS last updated on 12/Mar/18

	${let}^{'} s check for one pair of points$ $f (x) = \frac{x^{2}}{4} - \frac{x}{2} + \frac{5}{4}$ $f^{'} (x) = \frac{x - 1}{2} = k_{t a n g e n t s} = k_{t}$ $k_{n o r m a l s} = - \frac{2}{x - 1} = k_{n}$ $P = (\begin{matrix} 7 \\ 10 \end{matrix}); k_{t} = 3; k_{n} = - \frac{1}{3}$ $normal in P :$ $y = - \frac{x}{3} + d \Rightarrow 10 = - \frac{7}{3} + d \Rightarrow d = \frac{37}{3}$ $n_{P} : y = - \frac{x}{3} + \frac{37}{3}$ $now we search Q with Q \in p a r and$ $n_{Q} ∥ t_{P} (\Rightarrow n_{Q} ⊥ n_{P})$ $Q = (\begin{matrix} q \\ f (q) \end{matrix})$ $n_{Q} : y = k x + d with k = k_{n} = - \frac{2}{q - 1}$ $and k_{n} in Q = k_{t} in P \Rightarrow$ $\Rightarrow - \frac{2}{q - 1} = 3 \Rightarrow q = \frac{1}{3} \Rightarrow f (q) = \frac{10}{9}$ $Q = (\begin{matrix} \frac{1}{3} \\ \frac{10}{9} \end{matrix})$ $normal in Q :$ $y = 3 x + d \Rightarrow \frac{10}{9} = 1 + d \Rightarrow d = \frac{1}{9}$ $n_{Q} : y = 3 x + \frac{1}{9}$ $intersection of n_{P} and n_{Q} :$ $n_{P} : y = - \frac{x}{3} + \frac{37}{3}$ $n_{Q} : y = 3 x + \frac{1}{9}$ $3 x + \frac{1}{9} = - \frac{x}{3} + \frac{37}{3} \Rightarrow x = \frac{11}{3} \Rightarrow y = \frac{100}{9}$ $S_{n_{P} n_{Q}} = (\begin{matrix} \frac{11}{3} \\ \frac{100}{9} \end{matrix})$ $intersection point on ‘ ‘ {my}^{″} parabola :$ $y = x^{2} - 2 x + 5 = \frac{100}{9}$ $or on ‘ ‘ {book}^{'} s^{″} parabola :$ $y = x^{2} - 2 x + 3 = \frac{82}{9}$ $so the book is wrong$
	Commented by Tinkutara last updated on 12/Mar/18
	Thank you very much Sir! I got the answer. ��
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