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The shortest distance between the parabolas $y^2=x-1$ and $x^2=y-1$ is.

Attempt: The shortest distance is along the common normal of the two curves.

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    $\begingroup$ That's hardly an attempt, that's just an idea. An attempt would involve using it. An in general it's wrong, if you looked at $y=x^2$ and $y=x^2+1$, the only common normal would be the $y$-axis, but the distance along that is the maximum. $\endgroup$ – Henrik - stop hurting Monica Mar 28 '15 at 9:15
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    $\begingroup$ What mathematical resources can you use? For example, can you use partial derivatives of a function of two variables? $\endgroup$ – Rory Daulton Mar 28 '15 at 9:16
  • $\begingroup$ Henrik, yasir's statement is correct in the general case as long as there are no intersections and a minimum distance exists (which your example does not have) since the claim was not that the length of the common normal is always the minimum length but just that the minimum length is on a common normal. $\endgroup$ – Mint Mar 26 '18 at 4:32
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Hope a bit of calculus is allowed. Primes are with respect to x. Notice that they are inverse functions of each other, you can swap $ x,y$ to get to the second parabola. They are mirror images with respect to line $ x=y$. Required point should have this slope $y^{'} =1 $ for its tangent at point of tangency at ends of common normal.

Take the parabola with its symmetry axis coinciding with axis.

Differentiating $ 2 y y'= 1 , 2 y = 1, $ and the $x,y$ coordinates are

$$ (\dfrac54,\dfrac12)$$

and the other point of tangency is again swapped to

$$ (\dfrac12,\dfrac54); $$

Now use distance formula between the tangent points to get

$ d = \dfrac{3 \sqrt{2}}{4}. $

ParabolasSameSlope

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  • $\begingroup$ thanks. that was quite helpful. $\endgroup$ – yasir Mar 28 '15 at 14:52
  • $\begingroup$ You are welcome. $\endgroup$ – Narasimham Mar 28 '15 at 15:27
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    $\begingroup$ Could you explain why required point should have this slope $y′=1$ for its tangent at point of tangency at ends of common normal? $\endgroup$ – MathGeek Aug 28 '16 at 17:37
  • $\begingroup$ As shown in the sketch since the parabolas are symmetric with respect to $x=y, $ the common normal should have the same slope $=-1$ and the parallels should have same slope $=+1.$. $\endgroup$ – Narasimham Aug 28 '16 at 18:23
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    $\begingroup$ From the graph of any function its inverse fuction grapph can be obtained by mirroring about the straight line $x=y$. From points of maximum (or minimum) distance drop normals on to this line $x=y$. It is in fact the same normal line.The parallel lines have slope $m_1= +1$ and normal lines have slope $m_2= -1/m_1 = -1$. $\endgroup$ – Narasimham Mar 16 '18 at 11:28
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A point lying on the first parabola has coordinates $(1+u^2,u)$ while a point lying on the second parabola has coordinates $(v,1+v^2)$, hence the squared distance between them is given by: $$ d(u,v) = (1+u^2-v)^2 + (1+v^2-u)^2 $$ and the stationary points for such a function are given by the solutions of: $$ \frac{\partial d}{\partial u}= 4u(1+u^2-v)-2(1+v^2-u)=0,$$ $$ \frac{\partial d}{\partial v}= 4v(1+v^2-u)-2(1+u^2-v)=0,$$ hence they fulfill, by taking the difference between the two equations: $$ u(3+u+2u^2)=v(3+v+2v^2)\tag{1} $$ but since $\frac{d}{dt}\left(t(3+t+2t^2)\right) = 3+2t+6t^2$ has a negative discriminant we have that $t\to t(3+t+2t^2)$ is an injective function, hence $(1)$ implies $u=v$ and we have: $$ (4u-2)(1-u+u^2) = 0 \tag{2}$$ from which $u=v=\frac{1}{2}$ and: $$ \min_{u,v} {d(u,v)}^2 = 2\cdot\left(1+\frac{1}{4}-\frac{1}{2}\right)^2 = \frac{9}{8}.\tag{3}$$

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let the shortest distance between the parabolas $y = 1 + x^2$ and $x = 1 + y^2$ be given by $AB,$ where $A = (a , 1 + a^2), B = (1 + b^2, b)$ with both $a$ and $b$ positive. the tangent at $A$ has slope $2a$ and the one at $B$ has slope $\frac 1{2b}$ and the slope of $AB$ is $\frac{1+a^2 - b}{a-1-b^2}$

we need $$2a = \frac 1{2b} = \frac{1+b^2 - a}{1 + a^2 - b}$$ so that both tangents are parallel and orthogonal to $AB.$ this gives $$b = \frac 1 a, \, 2a = \frac{1 + \frac 1 {a^2} - a}{1 + a^2 - \frac 1 a} = \frac{a^2 + 1 - a^3}{a(a+a^3 -1)} \to 2a^3+2a^5-2a^2=a^2 + 1 - a^3$$ that is $$2a^5+3a^3 - 3a^2 - 1 = 0, a > 0. $$

numerically(ti-83 solver), i found $a = 0.9072$ to be the only solution.

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The two parabolas are clearly mirror images by the main diagonal ($x=y$), which diagonal they don't intersect (as $x^2-x+1>0$ for all real $x$). The shortest path then must be perpendicular to that diagonal, and the tangent to the parabolas parallel to the diagonal. The second parabola is the graph of the function $x\mapsto x^2+1$, and no complicated calculus is needed to see that the only point where its tangent is parallel to the diagonal is $(x,y)=(\frac12,\frac54)$. The distance to its mirror image $(\frac54,\frac12)$ is $\sqrt2\,\left|\frac54-\frac12\right|=\frac34\sqrt2$.

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    $\begingroup$ Really the same answer as that of Narashimham, as I discovered later. Please upvote that one (I've got plenty of points already). $\endgroup$ – Marc van Leeuwen Mar 28 '15 at 13:25

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