Find-the-shortest-distance-between-the-curve-y-2-x-1and-x-2-y-1-

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} = \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 \dots

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