Find the shortest distance between the curve y^2 = x−1and x^2 = y−1


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Question Number 134040 by benjo_mathlover last updated on 27/Feb/21

$Find the shortest distance$ $between the curve$ $y^{2} = x - 1 and x^{2} = y - 1$
Commented by benjo_mathlover last updated on 27/Feb/21

	Answered by EDWIN88 last updated on 27/Feb/21
	$(i) gradient of tangent line curve y^2 =x−1 at point A(a^2 +1, a) is m_1 = (1/(2(√(a^2 +1−1))))=(1/(2a)) then gradient of normal line is −2a⇒equation of normal line :y=−2a(x−a^2 −1)+a y=−2ax+2a^3 +3a (2)gradient of tangent line curve y=x^2 +1 at point B(b,b^2 +1) is m_2 = 2b then gradient of normal line is −(1/(2b)) ⇒equation of normal line : y=−(1/(2b))(x−b)+b^2 +1 y = −(1/(2b))x+ b^2 +(3/2) Thus the equation of normal line is same we get { ((2a=(1/(2b)) ; a=(1/(4b)))),((2a^3 +3a = b^2 +(3/2))) :} ⇔ (2/(64b^3 ))+(3/(4b)) = ((2b^2 +3)/2) ⇒((2+48b^2 )/(64b^3 )) = ((2b^2 +3)/2) ⇒ 2+48b^2 = 32b^3 (2b^2 +3) ⇒64b^5 +96b^3 −48b^2 −2=0 ⇒32b^5 +48b^3 −24b^2 −1=0 ⇒(2b−1)(16b^4 +8b^3 +28b^2 +2b+1)=0 { ((b=(1/2) ⇒B((1/2), (5/4)))),((a=(1/2)⇒A((5/4),(1/2)))) :} ⇒AB_(min) = (√(2((5/4)−(1/2))^2 ))=(3/4)(√2)$
	$(i) gradient of tangent line curve y^{2} = x - 1$ $at point A (a^{2} + 1, a) is m_{1} = \frac{1}{2 \sqrt{a^{2} + 1 - 1}} = \frac{1}{2 a} then$ $gradient of normal line is - 2 a \Rightarrow equation$ $of normal line : y = - 2 a (x - a^{2} - 1) + a$ $y = - 2 ax + {2 a}^{3} + 3 a$ $(2) gradient of tangent line curve y = x^{2} + 1$ $at point B (b, b^{2} + 1) is m_{2} = 2 b then$ $gradient of normal line is - \frac{1}{2 b} \Rightarrow equation$ $of normal line : y = - \frac{1}{2 b} (x - b) + b^{2} + 1$ $y = - \frac{1}{2 b} x + b^{2} + \frac{3}{2}$ $Thus the equation of normal line is same$ $we get {\begin{cases} 2 a = \frac{1}{2 b}; a = \frac{1}{4 b} \\ {2 a}^{3} + 3 a = b^{2} + \frac{3}{2} \end{cases}$ $\Leftrightarrow \frac{2}{{64 b}^{3}} + \frac{3}{4 b} = \frac{{2 b}^{2} + 3}{2} \Rightarrow \frac{2 + {48 b}^{2}}{{64 b}^{3}} = \frac{{2 b}^{2} + 3}{2}$ $\Rightarrow 2 + {48 b}^{2} = {32 b}^{3} ({2 b}^{2} + 3)$ $\Rightarrow {64 b}^{5} + {96 b}^{3} - {48 b}^{2} - 2 = 0$ $\Rightarrow {32 b}^{5} + {48 b}^{3} - {24 b}^{2} - 1 = 0$ $\Rightarrow (2 b - 1) ({16 b}^{4} + {8 b}^{3} + {28 b}^{2} + 2 b + 1) = 0$ ${\begin{cases} b = \frac{1}{2} \Rightarrow B (\frac{1}{2}, \frac{5}{4}) \\ a = \frac{1}{2} \Rightarrow A (\frac{5}{4}, \frac{1}{2}) \end{cases} \Rightarrow {AB}_{\min} = \sqrt{2 {(\frac{5}{4} - \frac{1}{2})}^{2}} = \frac{3}{4} \sqrt{2}$
	Answered by MJS_new last updated on 27/Feb/21

	$symmetry about y = x$ ${it}^{'} s very easy to solve . . .$
	Answered by benjo_mathlover last updated on 27/Feb/21

	$since both curve symmetry to line$ $y = x, gradient of normal line$ $curve y^{2} = x - 1 \Rightarrow - 2 \sqrt{x - 1} = - 1$ $\Rightarrow \sqrt{x - 1} = \frac{1}{2}; x = \frac{5}{4} \land y = \frac{1}{2}$ $we get point A (\frac{5}{4}, \frac{1}{2})$ $gradient of normal line curve$ $y = x^{2} + 1 \Rightarrow - \frac{1}{2 x} = - 1; x = \frac{1}{2} \land y = \frac{5}{4}$ $we get point B (\frac{1}{2}, \frac{5}{4}) .$ $So the shortest distance between$ $the given two curve is$ $AB = \sqrt{{(\frac{5}{4} - \frac{1}{2})}^{2} + {(\frac{1}{2} - \frac{5}{4})}^{2}}$ $= \sqrt{2 \times {(\frac{3}{4})}^{2}} = \frac{3 \sqrt{2}}{4}$
	Commented by benjo_mathlover last updated on 27/Feb/21

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