Transform parabola to be tangent to line in point and through other point












0












$begingroup$


Sorry for stupid question, but I give up. I've spent whole weekend to solve that and no results. Please help. I think I know the solution, but it doesn't work form me. So now I am not sure. I am not sure if I make some stupid mistake when simplifying equations or just my method is wrong? But solving that is very time consuming, so before next try I decided to first ask you if my method is OK, or should I change something?



I have line: $f_1(x)=ax$



I have point on that line: $ P _ { L x } $ ;$ P _ { L y }$



I also have second point that is not on the line (somewhere on the plane): $ P _ { x }$ ; $ P _ { y }$



I also already have parabola that I need to transform to make it tangent to line $f_1(x)$, and through the points $P_L $ and $P$



My parabola equation is: $f _ { 2 } ( x ) = b x ^ { 2 } + b x$



So it seems to be obvious I need to find some unknown variables $A$, $B$ and $C$. Like that:



$f _ { 2 } ( x ) = A b ( x - B ) ^ { 2 } + A b ( x - B ) + C$



So my next step is:
$f _ { 1 } left( x _ { 1 } right) = f _ { 2 } left( x _ { 1 } right)$



which is: $A b left( x _ { 1 } - B right) ^ { 2 } + A b left( x _ { 1 } - B right) + C = a x _ { 1 }$



Then: $f _ { 1 } ^ { prime } left( x _ { 1 } right) = f _ { 2 } ^ { prime } left( x _ { 1 } right)$



which is: $2 A b left( x _ { 1 } - B right) + A b = a$



So now I have four unknown: $x_1$, $A$, $B$, $C$, and two equations, so I need two more.



So I use point $P_L$:



$A b left( P _ { L x } - B right) ^ { 2 } + A c left( P _ { L y } - B right) + C = P _ { L y }$



And point $P$:



$A b left( P _ { x } - B right) ^ { 2 } + A c left( P _ { x } - B right) + C = P _ { y }$



So now I have four equations:



$A b left( x _ { 1 } - B right) ^ { 2 } + A b left( x _ { 1 } - B right) + C = a x _ { 1 }$



$2 A b left( x _ { 1 } - B right) + A b = a$



$A b left( P _ { L x } - B right) ^ { 2 } + A c left( P _ { L y } - B right) + C = P _ { L y }$



$A b left( P _ { x } - B right) ^ { 2 } + A c left( P _ { x } - B right) + C = P _ { y }$



And four unknown: $x_1$, $A$, $B$, $C$



Is that OK???



It doesn't work for me. I can't solve that. I end up with very complicated equations with a lot of square roots with many strange expressions and can't find the way to simplify them. I tried wolframalpha and www.mathway.com but they also say it's too complicated for them. So maybe I do something wrong?



For any help great thanks in advance.










share|cite|improve this question









$endgroup$












  • $begingroup$
    What transformations are you allowed to use?
    $endgroup$
    – amd
    Dec 3 '18 at 8:47










  • $begingroup$
    What do you mean „are you allowed”? I think I allowed all math that provide the solution
    $endgroup$
    – pajczur
    Dec 3 '18 at 8:50










  • $begingroup$
    Are you allowed to rotate the parabola? To apply a shear transformation to it? To reflect it about an arbitrary line? Perform an inversion in a circle or apply an arbitrary projective transformation on it? I suspect not. There’s likely some restricted set of transformations that you’re allowed to use to solve this problem. What are they?
    $endgroup$
    – amd
    Dec 3 '18 at 8:56












  • $begingroup$
    So actually I don’t know. Please notice my math skills are poor and additionally my English language is also poor. But I just asked if my method is properly? Or should I change something? Or is there any simpler method?
    $endgroup$
    – pajczur
    Dec 3 '18 at 9:10
















0












$begingroup$


Sorry for stupid question, but I give up. I've spent whole weekend to solve that and no results. Please help. I think I know the solution, but it doesn't work form me. So now I am not sure. I am not sure if I make some stupid mistake when simplifying equations or just my method is wrong? But solving that is very time consuming, so before next try I decided to first ask you if my method is OK, or should I change something?



I have line: $f_1(x)=ax$



I have point on that line: $ P _ { L x } $ ;$ P _ { L y }$



I also have second point that is not on the line (somewhere on the plane): $ P _ { x }$ ; $ P _ { y }$



I also already have parabola that I need to transform to make it tangent to line $f_1(x)$, and through the points $P_L $ and $P$



My parabola equation is: $f _ { 2 } ( x ) = b x ^ { 2 } + b x$



So it seems to be obvious I need to find some unknown variables $A$, $B$ and $C$. Like that:



$f _ { 2 } ( x ) = A b ( x - B ) ^ { 2 } + A b ( x - B ) + C$



So my next step is:
$f _ { 1 } left( x _ { 1 } right) = f _ { 2 } left( x _ { 1 } right)$



which is: $A b left( x _ { 1 } - B right) ^ { 2 } + A b left( x _ { 1 } - B right) + C = a x _ { 1 }$



Then: $f _ { 1 } ^ { prime } left( x _ { 1 } right) = f _ { 2 } ^ { prime } left( x _ { 1 } right)$



which is: $2 A b left( x _ { 1 } - B right) + A b = a$



So now I have four unknown: $x_1$, $A$, $B$, $C$, and two equations, so I need two more.



So I use point $P_L$:



$A b left( P _ { L x } - B right) ^ { 2 } + A c left( P _ { L y } - B right) + C = P _ { L y }$



And point $P$:



$A b left( P _ { x } - B right) ^ { 2 } + A c left( P _ { x } - B right) + C = P _ { y }$



So now I have four equations:



$A b left( x _ { 1 } - B right) ^ { 2 } + A b left( x _ { 1 } - B right) + C = a x _ { 1 }$



$2 A b left( x _ { 1 } - B right) + A b = a$



$A b left( P _ { L x } - B right) ^ { 2 } + A c left( P _ { L y } - B right) + C = P _ { L y }$



$A b left( P _ { x } - B right) ^ { 2 } + A c left( P _ { x } - B right) + C = P _ { y }$



And four unknown: $x_1$, $A$, $B$, $C$



Is that OK???



It doesn't work for me. I can't solve that. I end up with very complicated equations with a lot of square roots with many strange expressions and can't find the way to simplify them. I tried wolframalpha and www.mathway.com but they also say it's too complicated for them. So maybe I do something wrong?



For any help great thanks in advance.










share|cite|improve this question









$endgroup$












  • $begingroup$
    What transformations are you allowed to use?
    $endgroup$
    – amd
    Dec 3 '18 at 8:47










  • $begingroup$
    What do you mean „are you allowed”? I think I allowed all math that provide the solution
    $endgroup$
    – pajczur
    Dec 3 '18 at 8:50










  • $begingroup$
    Are you allowed to rotate the parabola? To apply a shear transformation to it? To reflect it about an arbitrary line? Perform an inversion in a circle or apply an arbitrary projective transformation on it? I suspect not. There’s likely some restricted set of transformations that you’re allowed to use to solve this problem. What are they?
    $endgroup$
    – amd
    Dec 3 '18 at 8:56












  • $begingroup$
    So actually I don’t know. Please notice my math skills are poor and additionally my English language is also poor. But I just asked if my method is properly? Or should I change something? Or is there any simpler method?
    $endgroup$
    – pajczur
    Dec 3 '18 at 9:10














0












0








0





$begingroup$


Sorry for stupid question, but I give up. I've spent whole weekend to solve that and no results. Please help. I think I know the solution, but it doesn't work form me. So now I am not sure. I am not sure if I make some stupid mistake when simplifying equations or just my method is wrong? But solving that is very time consuming, so before next try I decided to first ask you if my method is OK, or should I change something?



I have line: $f_1(x)=ax$



I have point on that line: $ P _ { L x } $ ;$ P _ { L y }$



I also have second point that is not on the line (somewhere on the plane): $ P _ { x }$ ; $ P _ { y }$



I also already have parabola that I need to transform to make it tangent to line $f_1(x)$, and through the points $P_L $ and $P$



My parabola equation is: $f _ { 2 } ( x ) = b x ^ { 2 } + b x$



So it seems to be obvious I need to find some unknown variables $A$, $B$ and $C$. Like that:



$f _ { 2 } ( x ) = A b ( x - B ) ^ { 2 } + A b ( x - B ) + C$



So my next step is:
$f _ { 1 } left( x _ { 1 } right) = f _ { 2 } left( x _ { 1 } right)$



which is: $A b left( x _ { 1 } - B right) ^ { 2 } + A b left( x _ { 1 } - B right) + C = a x _ { 1 }$



Then: $f _ { 1 } ^ { prime } left( x _ { 1 } right) = f _ { 2 } ^ { prime } left( x _ { 1 } right)$



which is: $2 A b left( x _ { 1 } - B right) + A b = a$



So now I have four unknown: $x_1$, $A$, $B$, $C$, and two equations, so I need two more.



So I use point $P_L$:



$A b left( P _ { L x } - B right) ^ { 2 } + A c left( P _ { L y } - B right) + C = P _ { L y }$



And point $P$:



$A b left( P _ { x } - B right) ^ { 2 } + A c left( P _ { x } - B right) + C = P _ { y }$



So now I have four equations:



$A b left( x _ { 1 } - B right) ^ { 2 } + A b left( x _ { 1 } - B right) + C = a x _ { 1 }$



$2 A b left( x _ { 1 } - B right) + A b = a$



$A b left( P _ { L x } - B right) ^ { 2 } + A c left( P _ { L y } - B right) + C = P _ { L y }$



$A b left( P _ { x } - B right) ^ { 2 } + A c left( P _ { x } - B right) + C = P _ { y }$



And four unknown: $x_1$, $A$, $B$, $C$



Is that OK???



It doesn't work for me. I can't solve that. I end up with very complicated equations with a lot of square roots with many strange expressions and can't find the way to simplify them. I tried wolframalpha and www.mathway.com but they also say it's too complicated for them. So maybe I do something wrong?



For any help great thanks in advance.










share|cite|improve this question









$endgroup$




Sorry for stupid question, but I give up. I've spent whole weekend to solve that and no results. Please help. I think I know the solution, but it doesn't work form me. So now I am not sure. I am not sure if I make some stupid mistake when simplifying equations or just my method is wrong? But solving that is very time consuming, so before next try I decided to first ask you if my method is OK, or should I change something?



I have line: $f_1(x)=ax$



I have point on that line: $ P _ { L x } $ ;$ P _ { L y }$



I also have second point that is not on the line (somewhere on the plane): $ P _ { x }$ ; $ P _ { y }$



I also already have parabola that I need to transform to make it tangent to line $f_1(x)$, and through the points $P_L $ and $P$



My parabola equation is: $f _ { 2 } ( x ) = b x ^ { 2 } + b x$



So it seems to be obvious I need to find some unknown variables $A$, $B$ and $C$. Like that:



$f _ { 2 } ( x ) = A b ( x - B ) ^ { 2 } + A b ( x - B ) + C$



So my next step is:
$f _ { 1 } left( x _ { 1 } right) = f _ { 2 } left( x _ { 1 } right)$



which is: $A b left( x _ { 1 } - B right) ^ { 2 } + A b left( x _ { 1 } - B right) + C = a x _ { 1 }$



Then: $f _ { 1 } ^ { prime } left( x _ { 1 } right) = f _ { 2 } ^ { prime } left( x _ { 1 } right)$



which is: $2 A b left( x _ { 1 } - B right) + A b = a$



So now I have four unknown: $x_1$, $A$, $B$, $C$, and two equations, so I need two more.



So I use point $P_L$:



$A b left( P _ { L x } - B right) ^ { 2 } + A c left( P _ { L y } - B right) + C = P _ { L y }$



And point $P$:



$A b left( P _ { x } - B right) ^ { 2 } + A c left( P _ { x } - B right) + C = P _ { y }$



So now I have four equations:



$A b left( x _ { 1 } - B right) ^ { 2 } + A b left( x _ { 1 } - B right) + C = a x _ { 1 }$



$2 A b left( x _ { 1 } - B right) + A b = a$



$A b left( P _ { L x } - B right) ^ { 2 } + A c left( P _ { L y } - B right) + C = P _ { L y }$



$A b left( P _ { x } - B right) ^ { 2 } + A c left( P _ { x } - B right) + C = P _ { y }$



And four unknown: $x_1$, $A$, $B$, $C$



Is that OK???



It doesn't work for me. I can't solve that. I end up with very complicated equations with a lot of square roots with many strange expressions and can't find the way to simplify them. I tried wolframalpha and www.mathway.com but they also say it's too complicated for them. So maybe I do something wrong?



For any help great thanks in advance.







conic-sections floating-point tangent-line-method






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 22:13









pajczurpajczur

854




854












  • $begingroup$
    What transformations are you allowed to use?
    $endgroup$
    – amd
    Dec 3 '18 at 8:47










  • $begingroup$
    What do you mean „are you allowed”? I think I allowed all math that provide the solution
    $endgroup$
    – pajczur
    Dec 3 '18 at 8:50










  • $begingroup$
    Are you allowed to rotate the parabola? To apply a shear transformation to it? To reflect it about an arbitrary line? Perform an inversion in a circle or apply an arbitrary projective transformation on it? I suspect not. There’s likely some restricted set of transformations that you’re allowed to use to solve this problem. What are they?
    $endgroup$
    – amd
    Dec 3 '18 at 8:56












  • $begingroup$
    So actually I don’t know. Please notice my math skills are poor and additionally my English language is also poor. But I just asked if my method is properly? Or should I change something? Or is there any simpler method?
    $endgroup$
    – pajczur
    Dec 3 '18 at 9:10


















  • $begingroup$
    What transformations are you allowed to use?
    $endgroup$
    – amd
    Dec 3 '18 at 8:47










  • $begingroup$
    What do you mean „are you allowed”? I think I allowed all math that provide the solution
    $endgroup$
    – pajczur
    Dec 3 '18 at 8:50










  • $begingroup$
    Are you allowed to rotate the parabola? To apply a shear transformation to it? To reflect it about an arbitrary line? Perform an inversion in a circle or apply an arbitrary projective transformation on it? I suspect not. There’s likely some restricted set of transformations that you’re allowed to use to solve this problem. What are they?
    $endgroup$
    – amd
    Dec 3 '18 at 8:56












  • $begingroup$
    So actually I don’t know. Please notice my math skills are poor and additionally my English language is also poor. But I just asked if my method is properly? Or should I change something? Or is there any simpler method?
    $endgroup$
    – pajczur
    Dec 3 '18 at 9:10
















$begingroup$
What transformations are you allowed to use?
$endgroup$
– amd
Dec 3 '18 at 8:47




$begingroup$
What transformations are you allowed to use?
$endgroup$
– amd
Dec 3 '18 at 8:47












$begingroup$
What do you mean „are you allowed”? I think I allowed all math that provide the solution
$endgroup$
– pajczur
Dec 3 '18 at 8:50




$begingroup$
What do you mean „are you allowed”? I think I allowed all math that provide the solution
$endgroup$
– pajczur
Dec 3 '18 at 8:50












$begingroup$
Are you allowed to rotate the parabola? To apply a shear transformation to it? To reflect it about an arbitrary line? Perform an inversion in a circle or apply an arbitrary projective transformation on it? I suspect not. There’s likely some restricted set of transformations that you’re allowed to use to solve this problem. What are they?
$endgroup$
– amd
Dec 3 '18 at 8:56






$begingroup$
Are you allowed to rotate the parabola? To apply a shear transformation to it? To reflect it about an arbitrary line? Perform an inversion in a circle or apply an arbitrary projective transformation on it? I suspect not. There’s likely some restricted set of transformations that you’re allowed to use to solve this problem. What are they?
$endgroup$
– amd
Dec 3 '18 at 8:56














$begingroup$
So actually I don’t know. Please notice my math skills are poor and additionally my English language is also poor. But I just asked if my method is properly? Or should I change something? Or is there any simpler method?
$endgroup$
– pajczur
Dec 3 '18 at 9:10




$begingroup$
So actually I don’t know. Please notice my math skills are poor and additionally my English language is also poor. But I just asked if my method is properly? Or should I change something? Or is there any simpler method?
$endgroup$
– pajczur
Dec 3 '18 at 9:10










1 Answer
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oldest

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$begingroup$

I suppose you want to find the equation of a parabola, having its axis parallel to the $y$-axis, passing through a given point $P$ and touching a given line (of equation $y=ax$) at a given point $P_L$.



If so, the solution is quite straightforward. A parabola of that kind has a an equation of the form $y=alpha x^2+beta x+gamma$ and passage through $P$ gives a first equation: $P_y=alpha P_x^2+beta P_x+gamma$. Passage through point $P_L=(P_{Lx},aP_{Lx})$ gives a second equation: $aP_{Lx}=alpha P_{Lx}^2+beta P_{Lx}+gamma$ and tangency at $P_L$ gives a third equation: $2alpha P_{Lx}+beta=a$.



You have then a system of three linear equations in three unknowns $alpha$, $beta$ and $gamma$.






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    $begingroup$

    I suppose you want to find the equation of a parabola, having its axis parallel to the $y$-axis, passing through a given point $P$ and touching a given line (of equation $y=ax$) at a given point $P_L$.



    If so, the solution is quite straightforward. A parabola of that kind has a an equation of the form $y=alpha x^2+beta x+gamma$ and passage through $P$ gives a first equation: $P_y=alpha P_x^2+beta P_x+gamma$. Passage through point $P_L=(P_{Lx},aP_{Lx})$ gives a second equation: $aP_{Lx}=alpha P_{Lx}^2+beta P_{Lx}+gamma$ and tangency at $P_L$ gives a third equation: $2alpha P_{Lx}+beta=a$.



    You have then a system of three linear equations in three unknowns $alpha$, $beta$ and $gamma$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I suppose you want to find the equation of a parabola, having its axis parallel to the $y$-axis, passing through a given point $P$ and touching a given line (of equation $y=ax$) at a given point $P_L$.



      If so, the solution is quite straightforward. A parabola of that kind has a an equation of the form $y=alpha x^2+beta x+gamma$ and passage through $P$ gives a first equation: $P_y=alpha P_x^2+beta P_x+gamma$. Passage through point $P_L=(P_{Lx},aP_{Lx})$ gives a second equation: $aP_{Lx}=alpha P_{Lx}^2+beta P_{Lx}+gamma$ and tangency at $P_L$ gives a third equation: $2alpha P_{Lx}+beta=a$.



      You have then a system of three linear equations in three unknowns $alpha$, $beta$ and $gamma$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I suppose you want to find the equation of a parabola, having its axis parallel to the $y$-axis, passing through a given point $P$ and touching a given line (of equation $y=ax$) at a given point $P_L$.



        If so, the solution is quite straightforward. A parabola of that kind has a an equation of the form $y=alpha x^2+beta x+gamma$ and passage through $P$ gives a first equation: $P_y=alpha P_x^2+beta P_x+gamma$. Passage through point $P_L=(P_{Lx},aP_{Lx})$ gives a second equation: $aP_{Lx}=alpha P_{Lx}^2+beta P_{Lx}+gamma$ and tangency at $P_L$ gives a third equation: $2alpha P_{Lx}+beta=a$.



        You have then a system of three linear equations in three unknowns $alpha$, $beta$ and $gamma$.






        share|cite|improve this answer









        $endgroup$



        I suppose you want to find the equation of a parabola, having its axis parallel to the $y$-axis, passing through a given point $P$ and touching a given line (of equation $y=ax$) at a given point $P_L$.



        If so, the solution is quite straightforward. A parabola of that kind has a an equation of the form $y=alpha x^2+beta x+gamma$ and passage through $P$ gives a first equation: $P_y=alpha P_x^2+beta P_x+gamma$. Passage through point $P_L=(P_{Lx},aP_{Lx})$ gives a second equation: $aP_{Lx}=alpha P_{Lx}^2+beta P_{Lx}+gamma$ and tangency at $P_L$ gives a third equation: $2alpha P_{Lx}+beta=a$.



        You have then a system of three linear equations in three unknowns $alpha$, $beta$ and $gamma$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 23:04









        AretinoAretino

        23.1k21443




        23.1k21443






























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