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Question Number 164178 by mr W last updated on 15/Jan/22

Commented by mr W last updated on 15/Jan/22

$t w o p o i n t s P, Q l i e o n t h e p a r a b o l a$ $w i t h P Q = b . f i n d t h e l o c u s o f i t s$ $m i d p o i n t M . f i n d t h e l o w e s t p o s s i b l e$ $p o s i t i o n o f M .$
	Answered by mindispower last updated on 15/Jan/22

	$P (a, a^{2}); Q (b, b^{2})$ $M = (\frac{a + b}{2}, \frac{a^{2} + b^{2}}{2})$ $P Q = l \Rightarrow {(b - a)}^{2} + {(b^{2} - a^{2})}^{2} = l^{2}$ ${(b - a)}^{2} (1 + {(b + a)}^{2}) = l^{2}$ $b - a = y$ $b + a = z$ $y^{2} (1 + z^{2}) = l^{2}$ $a^{2} + b^{2} = \frac{y^{2} + z^{2}}{2}$ $w e W a n t M i n (\frac{1}{8} (y^{2} + z^{2}) ∵ y^{2} (1 + z^{2}) = l^{2}$ $f (x, y, γ) = \frac{1}{8} y^{2} + z^{2} - γ (y^{2} (1 + z^{2}) - l^{2})$ $\partial_{y} f = 0 \Leftrightarrow \frac{1}{4} y - 2 y γ (1 + z^{2}) = 0$ $y = 0 o r γ = \frac{1}{8 (1 + z^{2})}$ $\partial z f = 0 \Leftrightarrow 0 \Rightarrow \frac{1}{4} z - 2 γ y^{2} z = 0$ $z (1 - 8 γ y^{2}) = 0$ $y^{2} (1 + z^{2}) = l^{2}$ $y = 0 \Rightarrow l = 0, z = 0 P = Q = O (0, 0)$ $l > 0, z = 0 \Rightarrow y^{2} = l^{2}, y (1 - 8 γ) = 0$ $y = b - a, 0 = a + b$ $b = - a = \frac{y}{2} M = \frac{a^{2} + b^{2}}{2} = \frac{l^{2}}{4}$ $γ = \frac{1}{8 (1 + z^{2})} = \frac{1}{8 y^{2}}$ $y^{2} = 1 + z^{2}$ $y^{4} = l^{2}$ $y^{2} = l$ $z^{2} = l - 1, l ⩾ 1$ $\frac{l}{4} ⩽ \frac{l^{2}}{4}$ $l = 0 m i n = 0$ $l \in [0, 1] m i n = \frac{l^{2}}{4}$ $l > 1 m i n = \frac{l}{4} i d i d q u i c k i w i l l t c h e k i t$
	Commented by mr W last updated on 15/Jan/22

	$t h a n k s s i r!$
	Commented by mindispower last updated on 15/Jan/22

	$p l e a s u r s i r n i c e d a y$
	Answered by mr W last updated on 15/Jan/22

	$p a r a b o l a y = \frac{x^{2}}{a} w i t h a = 1 h e r e$ $s a y P (p, \frac{p^{2}}{a}), Q (q, \frac{q^{2}}{a})$ $s a y M (u, v)$ $u = \frac{p + q}{2}$ $v = \frac{p^{2} + q^{2}}{2 a}$ $\Rightarrow p + q = 2 u$ $\Rightarrow p^{2} + q^{2} = 2 a v$ ${(p + q)}^{2} = p^{2} + q^{2} + 2 p q$ ${(p - q)}^{2} = p^{2} + q^{2} - 2 p q$ $\Rightarrow {(p - q)}^{2} + {(p + q)}^{2} = 2 (p^{2} + q^{2})$ $\Rightarrow {(p - q)}^{2} = 2 (2 a v) - {(2 u)}^{2} = 4 (a v - u^{2})$ ${P Q}^{2} = {(p - q)}^{2} + {(\frac{p^{2}}{a} - \frac{q^{2}}{a})}^{2} = b^{2}$ ${(p - q)}^{2} [1 + \frac{{(p + q)}^{2}}{a^{2}}] = b^{2}$ $4 (a v - u^{2}) [1 + \frac{{(2 u)}^{2}}{a^{2}}] = b^{2}$ $4 (a v - u^{2}) (1 + \frac{4 u^{2}}{a^{2}}) = b^{2}$ $v = \frac{u^{2}}{a} + \frac{{a b}^{2}}{4 (a^{2} + 4 u^{2})}$ $o r l o c u s o f p o i n t M i s$ $y = \frac{x^{2}}{a} + \frac{{a b}^{2}}{4 (a^{2} + 4 x^{2})}$ $y = \frac{4 x^{2} + a^{2}}{4 a} + \frac{{a b}^{2}}{4 (a^{2} + 4 x^{2})} - \frac{a}{4}$ $⩾ 2 \sqrt{\frac{4 x^{2} + a^{2}}{4 a} \times \frac{{a b}^{2}}{4 (a^{2} + 4 x^{2})}} - \frac{a}{4}$ $= \frac{b}{2} - \frac{a}{4}$ $i . e . y_{m i n} = \frac{b}{2} - \frac{a}{4} w h e n$ $\frac{4 x^{2} + a^{2}}{4 a} = \frac{{a b}^{2}}{4 (a^{2} + 4 x^{2})}$ $4 x^{2} + a^{2} = a b$ $x = \pm \frac{\sqrt{a (b - a)}}{2}$ $t h a t m e a n s t h e l o w e s t p o s i t i o n o f M$ $i s (\pm \frac{\sqrt{a (b - a)}}{2}, \frac{b}{2} - \frac{a}{4})$
	Commented by mr W last updated on 15/Jan/22

	Commented by Tawa11 last updated on 15/Jan/22

	$Great sir$
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