the-closest-distance-from-the-point-on-the-curve-y-x-3-1-to-the-curve-x-y-2-4-is-equal-to-

Question Number 135004 by bramlexs22 last updated on 09/Mar/21

the closest distance from the point on the curve y = x ^ 3-1 to the curve x = y ^2+ 4 is equal to

Commented by bramlexs22 last updated on 09/Mar/21

Commented by EDWIN88 last updated on 09/Mar/21

step (1) slope of tangent line curve y = x^{3} - 1 at point P (a, a^{3} - 1)

is m_{1} = {3 a}^{2} then slope of normal line is - \frac{1}{{3 a}^{2}}

the equation of normal line at point (a, a^{3} - 1)

is y = - \frac{1}{{3 a}^{2}} (x - a) + a^{3} - 1 \Rightarrow y = - \frac{x}{{3 a}^{2}} + \frac{1}{3 a} + a^{3} - 1

step (2) slope of tangent line curve y^{2} = x - 4 at point

Q (b^{2} + 4, b) is m_{2} = \frac{1}{2 b}, then slope of normal

line is - 2 b, and the equation of normal line

equal to y = - 2 b (x - b^{2} - 4) + b \Rightarrow y = - 2 bx + {2 b}^{3} + 9 b

step (3) the both of equation of normal line

is same, we get {\begin{cases} - \frac{1}{{3 a}^{2}} = - 2 b; a^{2} b = \frac{1}{6} \\ \frac{1}{3 a} + a^{3} - 1 = {2 b}^{3} + 9 b \end{cases}

solving for a and b, \Rightarrow \frac{1}{3 a} + a^{3} - 1 = 2 {(\frac{1}{{6 a}^{2}})}^{3} + 9 (\frac{1}{{6 a}^{2}})

\Rightarrow \frac{1}{3 a} + a^{3} - 1 = \frac{1}{{108 a}^{6}} + \frac{3}{{2 a}^{2}}

{36 a}^{5} + {108 a}^{9} - {108 a}^{6} = 1 + {162 a}^{4}

{108 a}^{9} - {108 a}^{6} + {36 a}^{5} - {162 a}^{4} - 1 = 0

we get {\begin{cases} a = 1.2067 \\ b = 0.1145 \end{cases}

we get a point P (1.2067; 0.7571) and

Q (4.0131; 0.1145)

therefore the closest distance is

∣ PQ ∣ = \sqrt{{(4.0131 - 1.2067)}^{2} + {(0.1145 - 0.7571)}^{2}}

∣ PQ ∣ \approx 2.87903

Commented by bramlexs22 last updated on 09/Mar/21

y = x^{3} - 1 to x = y^{2} + 4

Commented by bramlexs22 last updated on 10/Mar/21

Commented by mr W last updated on 09/Mar/21

E d w i n s i r :

p l e a s e c h e c k, i t h i n k e r r o r i s h e r e :

is same, we get {\begin{cases} - \frac{1}{{3 a}^{2}} = - 2 b; a^{2} b = \frac{1}{6} \\ \frac{1}{3 a} + a^{3} - 1 = {2 b}^{3} + 9 b \end{cases}

Commented by EDWIN88 last updated on 09/Mar/21

yes sir . thanks you sir

Answered by mr W last updated on 09/Mar/21

s a y p o i n t o n c u r v e y = x^{3} - 1 i s

P (p, p^{3} - 1)

p o i n t o n c u r v e x = y^{2} + 4 i s

Q (q^{2} + 4, q)

d = P Q

D = d^{2} = {(q^{2} + 4 - p)}^{2} + {(q - p^{3} + 1)}^{2}

\frac{\partial D}{\partial p} = - 2 (q^{2} + 4 - p) - 6 p^{2} (q - p^{3} + 1) = 0

\Rightarrow (q^{2} + 4 - p) + 3 p^{2} (q - p^{3} + 1) = 0 \dots (i)

\frac{\partial D}{\partial q} = 4 q (q^{2} + 4 - p) + 2 (q - p^{3} + 1) = 0

\Rightarrow 2 q (q^{2} + 4 - p) + (q - p^{3} + 1) = 0 \dots (i i)

\Rightarrow 2 q = \frac{1}{3 p^{2}} \Rightarrow q = \frac{1}{6 p^{2}}

p u t t h i s i n t o (i) :

\frac{1}{36 p^{4}} + 4 - p + 3 p^{2} (\frac{1}{6 p^{2}} - p^{3} + 1) = 0

\Rightarrow 108 p^{9} - 108 p^{6} + 36 p^{5} - 162 p^{4} - 1 = 0

\Rightarrow p \approx 1.2067

\Rightarrow d_{m i n} \approx 2.8791

Commented by mr W last updated on 09/Mar/21

Commented by bramlexs22 last updated on 09/Mar/21

thanks you sir