Mixing
Generalised method for finding the common normal of any two graphs in 2D
$begingroup$
I've always had trouble finding the common normals of two conic sections... In examinations, when I need to find the minimum distance between two graphs quickly, I find it very very difficult.
I've seen this question: Calculus - Finding the minimum vertical distance between graphs, but there are questions where this distance optimisation technique doesn't work well, due to the fact that it only finds the distance along the $y$-direction
For example:
Find the minimum radius of the circle touching the parabolas $x^2=y-1$ and $y^2=x-1$.
I'm quite sure that we have to find the minimum distance between them both, and then divide that distance by 2, because along the common normal, any circle drawn would have to be tangential to the two parabolas, and the minimum distance is also along the common normal. However, I'm still unable to find the actual common normal.
I'm asking this question NOT for the answer to the above mentioned example, but to learn, in general, how to find the common normal of any two general 2D graphs, which may even be $e^x$ and $y=sin(x)$, or anything for that matter...
Note that I'm asking if there is a method which works every time... If it doesn't exist, or can't be done, then a general method to find the distance between two general conics ($a_1x^2+b_1y^2+2h_1xy+2g_1x+2f_1y+c_1$ and $a_2x^2+b_2y^2+2h_2xy+2g_2x+2f_2y+c_2$) would also work...
calculus optimization analytic-geometry
asked Jan 30 at 1:26
AbhigyanCAbhigyanC
246315
$endgroup$
add a comment |
$begingroup$
I've always had trouble finding the common normals of two conic sections... In examinations, when I need to find the minimum distance between two graphs quickly, I find it very very difficult.
I've seen this question: Calculus - Finding the minimum vertical distance between graphs, but there are questions where this distance optimisation technique doesn't work well, due to the fact that it only finds the distance along the $y$-direction
For example:
Find the minimum radius of the circle touching the parabolas $x^2=y-1$ and $y^2=x-1$.
I'm quite sure that we have to find the minimum distance between them both, and then divide that distance by 2, because along the common normal, any circle drawn would have to be tangential to the two parabolas, and the minimum distance is also along the common normal. However, I'm still unable to find the actual common normal.
I'm asking this question NOT for the answer to the above mentioned example, but to learn, in general, how to find the common normal of any two general 2D graphs, which may even be $e^x$ and $y=sin(x)$, or anything for that matter...
Note that I'm asking if there is a method which works every time... If it doesn't exist, or can't be done, then a general method to find the distance between two general conics ($a_1x^2+b_1y^2+2h_1xy+2g_1x+2f_1y+c_1$ and $a_2x^2+b_2y^2+2h_2xy+2g_2x+2f_2y+c_2$) would also work...
calculus optimization analytic-geometry
asked Jan 30 at 1:26
AbhigyanCAbhigyanC
246315
$endgroup$
1
$begingroup$
In this case since the two curves are symmetric (the parabola on the x axis after a $pi /2 $ rotation counter clock wise gives you the second parabola). You can exploit the symmetry to obtain the slope of the normal (-1 ). After that it's easy work. For other no symmetric parabolas you'll have to assume a line $y = mx +c$, two points on both the parabolas then get 6 equations and solve them 2 from subbing in the points in the line, two from subbing in the points in the respective parabolas and two the normality condition.
$endgroup$
– Avnish Kabaj
Feb 6 at 12:08
add a comment |
$begingroup$
I've always had trouble finding the common normals of two conic sections... In examinations, when I need to find the minimum distance between two graphs quickly, I find it very very difficult.
I've seen this question: Calculus - Finding the minimum vertical distance between graphs, but there are questions where this distance optimisation technique doesn't work well, due to the fact that it only finds the distance along the $y$-direction
For example:
Find the minimum radius of the circle touching the parabolas $x^2=y-1$ and $y^2=x-1$.
I'm quite sure that we have to find the minimum distance between them both, and then divide that distance by 2, because along the common normal, any circle drawn would have to be tangential to the two parabolas, and the minimum distance is also along the common normal. However, I'm still unable to find the actual common normal.
I'm asking this question NOT for the answer to the above mentioned example, but to learn, in general, how to find the common normal of any two general 2D graphs, which may even be $e^x$ and $y=sin(x)$, or anything for that matter...
Note that I'm asking if there is a method which works every time... If it doesn't exist, or can't be done, then a general method to find the distance between two general conics ($a_1x^2+b_1y^2+2h_1xy+2g_1x+2f_1y+c_1$ and $a_2x^2+b_2y^2+2h_2xy+2g_2x+2f_2y+c_2$) would also work...
calculus optimization analytic-geometry
asked Jan 30 at 1:26
AbhigyanCAbhigyanC
246315
$endgroup$
I've always had trouble finding the common normals of two conic sections... In examinations, when I need to find the minimum distance between two graphs quickly, I find it very very difficult.
I've seen this question: Calculus - Finding the minimum vertical distance between graphs, but there are questions where this distance optimisation technique doesn't work well, due to the fact that it only finds the distance along the $y$-direction
For example:
Find the minimum radius of the circle touching the parabolas $x^2=y-1$ and $y^2=x-1$.
I'm quite sure that we have to find the minimum distance between them both, and then divide that distance by 2, because along the common normal, any circle drawn would have to be tangential to the two parabolas, and the minimum distance is also along the common normal. However, I'm still unable to find the actual common normal.
I'm asking this question NOT for the answer to the above mentioned example, but to learn, in general, how to find the common normal of any two general 2D graphs, which may even be $e^x$ and $y=sin(x)$, or anything for that matter...
Note that I'm asking if there is a method which works every time... If it doesn't exist, or can't be done, then a general method to find the distance between two general conics ($a_1x^2+b_1y^2+2h_1xy+2g_1x+2f_1y+c_1$ and $a_2x^2+b_2y^2+2h_2xy+2g_2x+2f_2y+c_2$) would also work...
calculus optimization analytic-geometry
calculus optimization analytic-geometry
asked Jan 30 at 1:26
AbhigyanCAbhigyanC
246315
asked Jan 30 at 1:26
AbhigyanCAbhigyanC
246315
edited Jan 30 at 1:32
AbhigyanC
asked Jan 30 at 1:26
AbhigyanCAbhigyanC
246315
asked Jan 30 at 1:26
AbhigyanCAbhigyanC
246315
asked Jan 30 at 1:26
AbhigyanCAbhigyanC
246315
246315
1
$begingroup$
In this case since the two curves are symmetric (the parabola on the x axis after a $pi /2 $ rotation counter clock wise gives you the second parabola). You can exploit the symmetry to obtain the slope of the normal (-1 ). After that it's easy work. For other no symmetric parabolas you'll have to assume a line $y = mx +c$, two points on both the parabolas then get 6 equations and solve them 2 from subbing in the points in the line, two from subbing in the points in the respective parabolas and two the normality condition.
$endgroup$
– Avnish Kabaj
Feb 6 at 12:08
add a comment |
1
$begingroup$
In this case since the two curves are symmetric (the parabola on the x axis after a $pi /2 $ rotation counter clock wise gives you the second parabola). You can exploit the symmetry to obtain the slope of the normal (-1 ). After that it's easy work. For other no symmetric parabolas you'll have to assume a line $y = mx +c$, two points on both the parabolas then get 6 equations and solve them 2 from subbing in the points in the line, two from subbing in the points in the respective parabolas and two the normality condition.
$endgroup$
– Avnish Kabaj
Feb 6 at 12:08
1
1
$begingroup$
In this case since the two curves are symmetric (the parabola on the x axis after a $pi /2 $ rotation counter clock wise gives you the second parabola). You can exploit the symmetry to obtain the slope of the normal (-1 ). After that it's easy work. For other no symmetric parabolas you'll have to assume a line $y = mx +c$, two points on both the parabolas then get 6 equations and solve them 2 from subbing in the points in the line, two from subbing in the points in the respective parabolas and two the normality condition.
$endgroup$
– Avnish Kabaj
Feb 6 at 12:08
$begingroup$
In this case since the two curves are symmetric (the parabola on the x axis after a $pi /2 $ rotation counter clock wise gives you the second parabola). You can exploit the symmetry to obtain the slope of the normal (-1 ). After that it's easy work. For other no symmetric parabolas you'll have to assume a line $y = mx +c$, two points on both the parabolas then get 6 equations and solve them 2 from subbing in the points in the line, two from subbing in the points in the respective parabolas and two the normality condition.
$endgroup$
– Avnish Kabaj
Feb 6 at 12:08
add a comment |
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$begingroup$
In this case since the two curves are symmetric (the parabola on the x axis after a $pi /2 $ rotation counter clock wise gives you the second parabola). You can exploit the symmetry to obtain the slope of the normal (-1 ). After that it's easy work. For other no symmetric parabolas you'll have to assume a line $y = mx +c$, two points on both the parabolas then get 6 equations and solve them 2 from subbing in the points in the line, two from subbing in the points in the respective parabolas and two the normality condition.
$endgroup$
– Avnish Kabaj
Feb 6 at 12:08