Mixing
Move parabola to make it tangent to lines
could anyone help me with something very easy (I hope).
I have two lines. For example:
$y = frac { x + 4.5 } { 6 }$
(which is green line on the graph)
and
$y = 1.2 x$
(which is red line on the graph)
And I also have parabola like that:
$y = x - 5 ( 0.5 x - 0.65 ) ^ { 2 }$
I need to find coordinates to move parabola to make it tangent to both lines and with shape of parabola untouched, and lines can't be moved.
Are there any general rules to make such things.
In the google I always find how to draw lines tangent to parabola and go through the point. But I can't figure it out how to make it in other direction
Graph now looks like that:
conic-sections tangent-line-method
edited Nov 22 '18 at 3:01
Andrei
11.5k21026
asked Nov 22 '18 at 2:36
pajczurpajczur
854
add a comment |
could anyone help me with something very easy (I hope).
I have two lines. For example:
$y = frac { x + 4.5 } { 6 }$
(which is green line on the graph)
and
$y = 1.2 x$
(which is red line on the graph)
And I also have parabola like that:
$y = x - 5 ( 0.5 x - 0.65 ) ^ { 2 }$
I need to find coordinates to move parabola to make it tangent to both lines and with shape of parabola untouched, and lines can't be moved.
Are there any general rules to make such things.
In the google I always find how to draw lines tangent to parabola and go through the point. But I can't figure it out how to make it in other direction
Graph now looks like that:
conic-sections tangent-line-method
edited Nov 22 '18 at 3:01
Andrei
11.5k21026
asked Nov 22 '18 at 2:36
pajczurpajczur
854
add a comment |
could anyone help me with something very easy (I hope).
I have two lines. For example:
$y = frac { x + 4.5 } { 6 }$
(which is green line on the graph)
and
$y = 1.2 x$
(which is red line on the graph)
And I also have parabola like that:
$y = x - 5 ( 0.5 x - 0.65 ) ^ { 2 }$
I need to find coordinates to move parabola to make it tangent to both lines and with shape of parabola untouched, and lines can't be moved.
Are there any general rules to make such things.
In the google I always find how to draw lines tangent to parabola and go through the point. But I can't figure it out how to make it in other direction
Graph now looks like that:
conic-sections tangent-line-method
edited Nov 22 '18 at 3:01
Andrei
11.5k21026
asked Nov 22 '18 at 2:36
pajczurpajczur
854
could anyone help me with something very easy (I hope).
I have two lines. For example:
$y = frac { x + 4.5 } { 6 }$
(which is green line on the graph)
and
$y = 1.2 x$
(which is red line on the graph)
And I also have parabola like that:
$y = x - 5 ( 0.5 x - 0.65 ) ^ { 2 }$
I need to find coordinates to move parabola to make it tangent to both lines and with shape of parabola untouched, and lines can't be moved.
Are there any general rules to make such things.
In the google I always find how to draw lines tangent to parabola and go through the point. But I can't figure it out how to make it in other direction
Graph now looks like that:
conic-sections tangent-line-method
conic-sections tangent-line-method
edited Nov 22 '18 at 3:01
Andrei
11.5k21026
asked Nov 22 '18 at 2:36
pajczurpajczur
854
edited Nov 22 '18 at 3:01
Andrei
11.5k21026
asked Nov 22 '18 at 2:36
pajczurpajczur
854
edited Nov 22 '18 at 3:01
Andrei
11.5k21026
edited Nov 22 '18 at 3:01
Andrei
11.5k21026
edited Nov 22 '18 at 3:01
Andrei
11.5k21026
11.5k21026
asked Nov 22 '18 at 2:36
pajczurpajczur
854
asked Nov 22 '18 at 2:36
pajczurpajczur
854
asked Nov 22 '18 at 2:36
pajczurpajczur
854
854
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You can move the parabola left-right by adding an offset to $x$ and you can move it up-down adding an offset to $y$. Let's solve the problem in the more general case. The original parabola is $y=ax^2+bx+c$. The new equation for the parabola will be $$y=a(x-x_0)^2+b(x-x_0)+c+y_0$$
We will need the derivative with respect to $x$ for the tangents, so $$y'=2a(x-x_0)+b$$
Now the equations of the lines are $$y=m_1x+n_1\y=m_2x+n_2$$
When the parabola is tangent to a line, it means that they intersect at one point, and that the slopes at that points are identical. So say we have the tangent point with line 1 at $x_1$ and with line 2 at $x_2$, so you get the following system of equations:
$$begin{align}a(x_1-x_0)^2+b(x_1-x_0)+c+y_0&=m_1x_1+n_1\
a(x_2-x_0)^2+b(x_2-x_0)+c+y_0&=m_2x_2+n_2\
2a(x_1-x_0)+b&=m_1\2a(x_2-x_0)+b&=m_2
end{align}$$
You have a system with four equations, with four unknowns $(x_0,y_0,x_1,x_2)$.
answered Nov 22 '18 at 3:15
AndreiAndrei
11.5k21026
Great thanks. I also have other question, about rotateing parabola round some point, but keeping tangent to two lines. But I think I should create separate subject. But one more, great thanks for your help
– pajczur
Nov 22 '18 at 8:03
Hey, sorry but there is something wrong: you told on the end there are equations with four unknowns (x0,y0,x1,y1), but in equations I can't find y1, and among unknowns I can't find x2 which are in equations. So which part is wrong?
– pajczur
Nov 22 '18 at 8:50
$(x_0,y_0,x_1,x_2)$ is the correct way. You can then find $y_1=m_1x_1+n_1$ and similarly for $y_2$. I've modified my answer.
– Andrei
Nov 22 '18 at 12:04
add a comment |
You can shift the graph of any function horizontally by a distance of $Delta x$ by replacing $x$ with $(x-Delta x)$, and vertically by simply adding $Delta y$ to its value. If the curve is given implicitly by $F(x,y)$, the latter corresponds to replacing $y$ by $(y-Delta y)$. In the same vein as Andrei’s answer, then, the problem is to find these offsets for your parabola. In contrast to that answer, we’ll work with the original parabola.
First, construct tangents to the parabola that are parallel to the two given lines. If the parabola is given by an equation of the form $y=f(x)$ (i.e., it’s the graph of $f$), an easy way to do this is to find the points at which the derivative of $f$ is equal to the two slopes. This is a matter of solving two very simple equations for $x$ and plugging those values into $f$. Now, find the intersection point $P$ of the two lines you’ve constructed, and the intersection $Q$ of the original two lines. The offset that you need is then simply $(Delta x,Delta y) = Q-P$. If you work in homogeneous coordinates, these computations can be done directly, without having to solve any other equations.
For your problem, we have $f(x) = x-5(0.5x-0.65)^2$, so $f'(x) = 1-5(0.5x-0.65)$. The slope of the first line is $1/6$, for which $x=49/30$ (assuming that those decimal fractions in $f$ are exact) and $y=f(49/30)=269/180$. For the second line the slope is $1.2$, giving $x=61/50$ and $y=303/250$. I assume that you know how to construct equations of the tangents through these points and finish the computation, but I’ll go through it with homogeneous coordinates to illustrate that method.
The line through two points is given by their cross product, as is the intersection of two lines. Also, if a line has slope $m$, then the point at infinity $(1,m,0)$ lies on that line. So, our two constructed tangent lines are $$(49/30,269,180,1)times(1,1/6,0)=(-1/6,1,-11/9),$$ i.e., $y=x/6+11/9$, and $$(61/50,303/250,1)times(1,12/10,0)=(-6/5,1,63/250).$$ Their intersection is therefore $$left(-frac16,1,-frac{11}9right)timesleft(-frac65,1,frac{63}{250}right) approx (1.474,1.509,1.033),$$ which we dehomogenize by dividing through by the last coordinate, yielding $Papprox(1.427,1.460)$. Intersecting the target lines, we have $$left(frac16,-1,frac{45}{60}right)timesleft(frac{12}{10},-1,0right) approx (0.750,0.900,1.033),$$ or $Qapprox (0.726,0.871)$, from which $Delta x approx -0.70$, $Delta yapprox -0.59$. The shifted parabola is therefore (to two decimal places) $$y = f(x+0.70)-0.59 = (x+0.70)-5(0.5(x+0.70)-0.65)^2-0.59.$$
answered Nov 24 '18 at 0:55
amdamd
29.4k21050
Hello great thanks for answer. But I have impression like now after your help I know little bit less. Is it really f’(x)=1 - 5(0.5x + 0.65). I was sure it should be f’(x)=1 - 2*5(0.5x + 0.65). But now I am not sure because your solution also works. So I am in great big black hole :) What’s wrong?
– pajczur
Nov 24 '18 at 3:11
@pajczur You forgot to multiply by $0.5$ for the derivative of the parenthesized expression. You might want to review the chain rule.
– amd
Nov 24 '18 at 3:12
Yes chain rule would be great. What should I multiply by $0.5$ ? Exactly which part?
– pajczur
Nov 24 '18 at 3:17
Ok sorry my fault, I’ve found it thanks
– pajczur
Nov 24 '18 at 4:02
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008672%2fmove-parabola-to-make-it-tangent-to-lines%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can move the parabola left-right by adding an offset to $x$ and you can move it up-down adding an offset to $y$. Let's solve the problem in the more general case. The original parabola is $y=ax^2+bx+c$. The new equation for the parabola will be $$y=a(x-x_0)^2+b(x-x_0)+c+y_0$$
We will need the derivative with respect to $x$ for the tangents, so $$y'=2a(x-x_0)+b$$
Now the equations of the lines are $$y=m_1x+n_1\y=m_2x+n_2$$
When the parabola is tangent to a line, it means that they intersect at one point, and that the slopes at that points are identical. So say we have the tangent point with line 1 at $x_1$ and with line 2 at $x_2$, so you get the following system of equations:
$$begin{align}a(x_1-x_0)^2+b(x_1-x_0)+c+y_0&=m_1x_1+n_1\
a(x_2-x_0)^2+b(x_2-x_0)+c+y_0&=m_2x_2+n_2\
2a(x_1-x_0)+b&=m_1\2a(x_2-x_0)+b&=m_2
end{align}$$
You have a system with four equations, with four unknowns $(x_0,y_0,x_1,x_2)$.
answered Nov 22 '18 at 3:15
AndreiAndrei
11.5k21026
Great thanks. I also have other question, about rotateing parabola round some point, but keeping tangent to two lines. But I think I should create separate subject. But one more, great thanks for your help
– pajczur
Nov 22 '18 at 8:03
Hey, sorry but there is something wrong: you told on the end there are equations with four unknowns (x0,y0,x1,y1), but in equations I can't find y1, and among unknowns I can't find x2 which are in equations. So which part is wrong?
– pajczur
Nov 22 '18 at 8:50
$(x_0,y_0,x_1,x_2)$ is the correct way. You can then find $y_1=m_1x_1+n_1$ and similarly for $y_2$. I've modified my answer.
– Andrei
Nov 22 '18 at 12:04
add a comment |
You can move the parabola left-right by adding an offset to $x$ and you can move it up-down adding an offset to $y$. Let's solve the problem in the more general case. The original parabola is $y=ax^2+bx+c$. The new equation for the parabola will be $$y=a(x-x_0)^2+b(x-x_0)+c+y_0$$
We will need the derivative with respect to $x$ for the tangents, so $$y'=2a(x-x_0)+b$$
Now the equations of the lines are $$y=m_1x+n_1\y=m_2x+n_2$$
When the parabola is tangent to a line, it means that they intersect at one point, and that the slopes at that points are identical. So say we have the tangent point with line 1 at $x_1$ and with line 2 at $x_2$, so you get the following system of equations:
$$begin{align}a(x_1-x_0)^2+b(x_1-x_0)+c+y_0&=m_1x_1+n_1\
a(x_2-x_0)^2+b(x_2-x_0)+c+y_0&=m_2x_2+n_2\
2a(x_1-x_0)+b&=m_1\2a(x_2-x_0)+b&=m_2
end{align}$$
You have a system with four equations, with four unknowns $(x_0,y_0,x_1,x_2)$.
answered Nov 22 '18 at 3:15
AndreiAndrei
11.5k21026
Great thanks. I also have other question, about rotateing parabola round some point, but keeping tangent to two lines. But I think I should create separate subject. But one more, great thanks for your help
– pajczur
Nov 22 '18 at 8:03
Hey, sorry but there is something wrong: you told on the end there are equations with four unknowns (x0,y0,x1,y1), but in equations I can't find y1, and among unknowns I can't find x2 which are in equations. So which part is wrong?
– pajczur
Nov 22 '18 at 8:50
$(x_0,y_0,x_1,x_2)$ is the correct way. You can then find $y_1=m_1x_1+n_1$ and similarly for $y_2$. I've modified my answer.
– Andrei
Nov 22 '18 at 12:04
add a comment |
You can move the parabola left-right by adding an offset to $x$ and you can move it up-down adding an offset to $y$. Let's solve the problem in the more general case. The original parabola is $y=ax^2+bx+c$. The new equation for the parabola will be $$y=a(x-x_0)^2+b(x-x_0)+c+y_0$$
We will need the derivative with respect to $x$ for the tangents, so $$y'=2a(x-x_0)+b$$
Now the equations of the lines are $$y=m_1x+n_1\y=m_2x+n_2$$
When the parabola is tangent to a line, it means that they intersect at one point, and that the slopes at that points are identical. So say we have the tangent point with line 1 at $x_1$ and with line 2 at $x_2$, so you get the following system of equations:
$$begin{align}a(x_1-x_0)^2+b(x_1-x_0)+c+y_0&=m_1x_1+n_1\
a(x_2-x_0)^2+b(x_2-x_0)+c+y_0&=m_2x_2+n_2\
2a(x_1-x_0)+b&=m_1\2a(x_2-x_0)+b&=m_2
end{align}$$
You have a system with four equations, with four unknowns $(x_0,y_0,x_1,x_2)$.
answered Nov 22 '18 at 3:15
AndreiAndrei
11.5k21026
You can move the parabola left-right by adding an offset to $x$ and you can move it up-down adding an offset to $y$. Let's solve the problem in the more general case. The original parabola is $y=ax^2+bx+c$. The new equation for the parabola will be $$y=a(x-x_0)^2+b(x-x_0)+c+y_0$$
We will need the derivative with respect to $x$ for the tangents, so $$y'=2a(x-x_0)+b$$
Now the equations of the lines are $$y=m_1x+n_1\y=m_2x+n_2$$
When the parabola is tangent to a line, it means that they intersect at one point, and that the slopes at that points are identical. So say we have the tangent point with line 1 at $x_1$ and with line 2 at $x_2$, so you get the following system of equations:
$$begin{align}a(x_1-x_0)^2+b(x_1-x_0)+c+y_0&=m_1x_1+n_1\
a(x_2-x_0)^2+b(x_2-x_0)+c+y_0&=m_2x_2+n_2\
2a(x_1-x_0)+b&=m_1\2a(x_2-x_0)+b&=m_2
end{align}$$
You have a system with four equations, with four unknowns $(x_0,y_0,x_1,x_2)$.
answered Nov 22 '18 at 3:15
AndreiAndrei
11.5k21026
edited Nov 22 '18 at 12:04
answered Nov 22 '18 at 3:15
AndreiAndrei
11.5k21026
answered Nov 22 '18 at 3:15
AndreiAndrei
11.5k21026
answered Nov 22 '18 at 3:15
AndreiAndrei
11.5k21026
11.5k21026
Great thanks. I also have other question, about rotateing parabola round some point, but keeping tangent to two lines. But I think I should create separate subject. But one more, great thanks for your help
– pajczur
Nov 22 '18 at 8:03
Hey, sorry but there is something wrong: you told on the end there are equations with four unknowns (x0,y0,x1,y1), but in equations I can't find y1, and among unknowns I can't find x2 which are in equations. So which part is wrong?
– pajczur
Nov 22 '18 at 8:50
$(x_0,y_0,x_1,x_2)$ is the correct way. You can then find $y_1=m_1x_1+n_1$ and similarly for $y_2$. I've modified my answer.
– Andrei
Nov 22 '18 at 12:04
add a comment |
Great thanks. I also have other question, about rotateing parabola round some point, but keeping tangent to two lines. But I think I should create separate subject. But one more, great thanks for your help
– pajczur
Nov 22 '18 at 8:03
Hey, sorry but there is something wrong: you told on the end there are equations with four unknowns (x0,y0,x1,y1), but in equations I can't find y1, and among unknowns I can't find x2 which are in equations. So which part is wrong?
– pajczur
Nov 22 '18 at 8:50
$(x_0,y_0,x_1,x_2)$ is the correct way. You can then find $y_1=m_1x_1+n_1$ and similarly for $y_2$. I've modified my answer.
– Andrei
Nov 22 '18 at 12:04
Great thanks. I also have other question, about rotateing parabola round some point, but keeping tangent to two lines. But I think I should create separate subject. But one more, great thanks for your help
– pajczur
Nov 22 '18 at 8:03
Great thanks. I also have other question, about rotateing parabola round some point, but keeping tangent to two lines. But I think I should create separate subject. But one more, great thanks for your help
– pajczur
Nov 22 '18 at 8:03
Hey, sorry but there is something wrong: you told on the end there are equations with four unknowns (x0,y0,x1,y1), but in equations I can't find y1, and among unknowns I can't find x2 which are in equations. So which part is wrong?
– pajczur
Nov 22 '18 at 8:50
Hey, sorry but there is something wrong: you told on the end there are equations with four unknowns (x0,y0,x1,y1), but in equations I can't find y1, and among unknowns I can't find x2 which are in equations. So which part is wrong?
– pajczur
Nov 22 '18 at 8:50
$(x_0,y_0,x_1,x_2)$ is the correct way. You can then find $y_1=m_1x_1+n_1$ and similarly for $y_2$. I've modified my answer.
– Andrei
Nov 22 '18 at 12:04
$(x_0,y_0,x_1,x_2)$ is the correct way. You can then find $y_1=m_1x_1+n_1$ and similarly for $y_2$. I've modified my answer.
– Andrei
Nov 22 '18 at 12:04
add a comment |
You can shift the graph of any function horizontally by a distance of $Delta x$ by replacing $x$ with $(x-Delta x)$, and vertically by simply adding $Delta y$ to its value. If the curve is given implicitly by $F(x,y)$, the latter corresponds to replacing $y$ by $(y-Delta y)$. In the same vein as Andrei’s answer, then, the problem is to find these offsets for your parabola. In contrast to that answer, we’ll work with the original parabola.
First, construct tangents to the parabola that are parallel to the two given lines. If the parabola is given by an equation of the form $y=f(x)$ (i.e., it’s the graph of $f$), an easy way to do this is to find the points at which the derivative of $f$ is equal to the two slopes. This is a matter of solving two very simple equations for $x$ and plugging those values into $f$. Now, find the intersection point $P$ of the two lines you’ve constructed, and the intersection $Q$ of the original two lines. The offset that you need is then simply $(Delta x,Delta y) = Q-P$. If you work in homogeneous coordinates, these computations can be done directly, without having to solve any other equations.
For your problem, we have $f(x) = x-5(0.5x-0.65)^2$, so $f'(x) = 1-5(0.5x-0.65)$. The slope of the first line is $1/6$, for which $x=49/30$ (assuming that those decimal fractions in $f$ are exact) and $y=f(49/30)=269/180$. For the second line the slope is $1.2$, giving $x=61/50$ and $y=303/250$. I assume that you know how to construct equations of the tangents through these points and finish the computation, but I’ll go through it with homogeneous coordinates to illustrate that method.
The line through two points is given by their cross product, as is the intersection of two lines. Also, if a line has slope $m$, then the point at infinity $(1,m,0)$ lies on that line. So, our two constructed tangent lines are $$(49/30,269,180,1)times(1,1/6,0)=(-1/6,1,-11/9),$$ i.e., $y=x/6+11/9$, and $$(61/50,303/250,1)times(1,12/10,0)=(-6/5,1,63/250).$$ Their intersection is therefore $$left(-frac16,1,-frac{11}9right)timesleft(-frac65,1,frac{63}{250}right) approx (1.474,1.509,1.033),$$ which we dehomogenize by dividing through by the last coordinate, yielding $Papprox(1.427,1.460)$. Intersecting the target lines, we have $$left(frac16,-1,frac{45}{60}right)timesleft(frac{12}{10},-1,0right) approx (0.750,0.900,1.033),$$ or $Qapprox (0.726,0.871)$, from which $Delta x approx -0.70$, $Delta yapprox -0.59$. The shifted parabola is therefore (to two decimal places) $$y = f(x+0.70)-0.59 = (x+0.70)-5(0.5(x+0.70)-0.65)^2-0.59.$$
answered Nov 24 '18 at 0:55
amdamd
29.4k21050
Hello great thanks for answer. But I have impression like now after your help I know little bit less. Is it really f’(x)=1 - 5(0.5x + 0.65). I was sure it should be f’(x)=1 - 2*5(0.5x + 0.65). But now I am not sure because your solution also works. So I am in great big black hole :) What’s wrong?
– pajczur
Nov 24 '18 at 3:11
@pajczur You forgot to multiply by $0.5$ for the derivative of the parenthesized expression. You might want to review the chain rule.
– amd
Nov 24 '18 at 3:12
Yes chain rule would be great. What should I multiply by $0.5$ ? Exactly which part?
– pajczur
Nov 24 '18 at 3:17
Ok sorry my fault, I’ve found it thanks
– pajczur
Nov 24 '18 at 4:02
add a comment |
You can shift the graph of any function horizontally by a distance of $Delta x$ by replacing $x$ with $(x-Delta x)$, and vertically by simply adding $Delta y$ to its value. If the curve is given implicitly by $F(x,y)$, the latter corresponds to replacing $y$ by $(y-Delta y)$. In the same vein as Andrei’s answer, then, the problem is to find these offsets for your parabola. In contrast to that answer, we’ll work with the original parabola.
First, construct tangents to the parabola that are parallel to the two given lines. If the parabola is given by an equation of the form $y=f(x)$ (i.e., it’s the graph of $f$), an easy way to do this is to find the points at which the derivative of $f$ is equal to the two slopes. This is a matter of solving two very simple equations for $x$ and plugging those values into $f$. Now, find the intersection point $P$ of the two lines you’ve constructed, and the intersection $Q$ of the original two lines. The offset that you need is then simply $(Delta x,Delta y) = Q-P$. If you work in homogeneous coordinates, these computations can be done directly, without having to solve any other equations.
For your problem, we have $f(x) = x-5(0.5x-0.65)^2$, so $f'(x) = 1-5(0.5x-0.65)$. The slope of the first line is $1/6$, for which $x=49/30$ (assuming that those decimal fractions in $f$ are exact) and $y=f(49/30)=269/180$. For the second line the slope is $1.2$, giving $x=61/50$ and $y=303/250$. I assume that you know how to construct equations of the tangents through these points and finish the computation, but I’ll go through it with homogeneous coordinates to illustrate that method.
The line through two points is given by their cross product, as is the intersection of two lines. Also, if a line has slope $m$, then the point at infinity $(1,m,0)$ lies on that line. So, our two constructed tangent lines are $$(49/30,269,180,1)times(1,1/6,0)=(-1/6,1,-11/9),$$ i.e., $y=x/6+11/9$, and $$(61/50,303/250,1)times(1,12/10,0)=(-6/5,1,63/250).$$ Their intersection is therefore $$left(-frac16,1,-frac{11}9right)timesleft(-frac65,1,frac{63}{250}right) approx (1.474,1.509,1.033),$$ which we dehomogenize by dividing through by the last coordinate, yielding $Papprox(1.427,1.460)$. Intersecting the target lines, we have $$left(frac16,-1,frac{45}{60}right)timesleft(frac{12}{10},-1,0right) approx (0.750,0.900,1.033),$$ or $Qapprox (0.726,0.871)$, from which $Delta x approx -0.70$, $Delta yapprox -0.59$. The shifted parabola is therefore (to two decimal places) $$y = f(x+0.70)-0.59 = (x+0.70)-5(0.5(x+0.70)-0.65)^2-0.59.$$
answered Nov 24 '18 at 0:55
amdamd
29.4k21050
Hello great thanks for answer. But I have impression like now after your help I know little bit less. Is it really f’(x)=1 - 5(0.5x + 0.65). I was sure it should be f’(x)=1 - 2*5(0.5x + 0.65). But now I am not sure because your solution also works. So I am in great big black hole :) What’s wrong?
– pajczur
Nov 24 '18 at 3:11
@pajczur You forgot to multiply by $0.5$ for the derivative of the parenthesized expression. You might want to review the chain rule.
– amd
Nov 24 '18 at 3:12
Yes chain rule would be great. What should I multiply by $0.5$ ? Exactly which part?
– pajczur
Nov 24 '18 at 3:17
Ok sorry my fault, I’ve found it thanks
– pajczur
Nov 24 '18 at 4:02
add a comment |
You can shift the graph of any function horizontally by a distance of $Delta x$ by replacing $x$ with $(x-Delta x)$, and vertically by simply adding $Delta y$ to its value. If the curve is given implicitly by $F(x,y)$, the latter corresponds to replacing $y$ by $(y-Delta y)$. In the same vein as Andrei’s answer, then, the problem is to find these offsets for your parabola. In contrast to that answer, we’ll work with the original parabola.
First, construct tangents to the parabola that are parallel to the two given lines. If the parabola is given by an equation of the form $y=f(x)$ (i.e., it’s the graph of $f$), an easy way to do this is to find the points at which the derivative of $f$ is equal to the two slopes. This is a matter of solving two very simple equations for $x$ and plugging those values into $f$. Now, find the intersection point $P$ of the two lines you’ve constructed, and the intersection $Q$ of the original two lines. The offset that you need is then simply $(Delta x,Delta y) = Q-P$. If you work in homogeneous coordinates, these computations can be done directly, without having to solve any other equations.
For your problem, we have $f(x) = x-5(0.5x-0.65)^2$, so $f'(x) = 1-5(0.5x-0.65)$. The slope of the first line is $1/6$, for which $x=49/30$ (assuming that those decimal fractions in $f$ are exact) and $y=f(49/30)=269/180$. For the second line the slope is $1.2$, giving $x=61/50$ and $y=303/250$. I assume that you know how to construct equations of the tangents through these points and finish the computation, but I’ll go through it with homogeneous coordinates to illustrate that method.
The line through two points is given by their cross product, as is the intersection of two lines. Also, if a line has slope $m$, then the point at infinity $(1,m,0)$ lies on that line. So, our two constructed tangent lines are $$(49/30,269,180,1)times(1,1/6,0)=(-1/6,1,-11/9),$$ i.e., $y=x/6+11/9$, and $$(61/50,303/250,1)times(1,12/10,0)=(-6/5,1,63/250).$$ Their intersection is therefore $$left(-frac16,1,-frac{11}9right)timesleft(-frac65,1,frac{63}{250}right) approx (1.474,1.509,1.033),$$ which we dehomogenize by dividing through by the last coordinate, yielding $Papprox(1.427,1.460)$. Intersecting the target lines, we have $$left(frac16,-1,frac{45}{60}right)timesleft(frac{12}{10},-1,0right) approx (0.750,0.900,1.033),$$ or $Qapprox (0.726,0.871)$, from which $Delta x approx -0.70$, $Delta yapprox -0.59$. The shifted parabola is therefore (to two decimal places) $$y = f(x+0.70)-0.59 = (x+0.70)-5(0.5(x+0.70)-0.65)^2-0.59.$$
answered Nov 24 '18 at 0:55
amdamd
29.4k21050
You can shift the graph of any function horizontally by a distance of $Delta x$ by replacing $x$ with $(x-Delta x)$, and vertically by simply adding $Delta y$ to its value. If the curve is given implicitly by $F(x,y)$, the latter corresponds to replacing $y$ by $(y-Delta y)$. In the same vein as Andrei’s answer, then, the problem is to find these offsets for your parabola. In contrast to that answer, we’ll work with the original parabola.
First, construct tangents to the parabola that are parallel to the two given lines. If the parabola is given by an equation of the form $y=f(x)$ (i.e., it’s the graph of $f$), an easy way to do this is to find the points at which the derivative of $f$ is equal to the two slopes. This is a matter of solving two very simple equations for $x$ and plugging those values into $f$. Now, find the intersection point $P$ of the two lines you’ve constructed, and the intersection $Q$ of the original two lines. The offset that you need is then simply $(Delta x,Delta y) = Q-P$. If you work in homogeneous coordinates, these computations can be done directly, without having to solve any other equations.
For your problem, we have $f(x) = x-5(0.5x-0.65)^2$, so $f'(x) = 1-5(0.5x-0.65)$. The slope of the first line is $1/6$, for which $x=49/30$ (assuming that those decimal fractions in $f$ are exact) and $y=f(49/30)=269/180$. For the second line the slope is $1.2$, giving $x=61/50$ and $y=303/250$. I assume that you know how to construct equations of the tangents through these points and finish the computation, but I’ll go through it with homogeneous coordinates to illustrate that method.
The line through two points is given by their cross product, as is the intersection of two lines. Also, if a line has slope $m$, then the point at infinity $(1,m,0)$ lies on that line. So, our two constructed tangent lines are $$(49/30,269,180,1)times(1,1/6,0)=(-1/6,1,-11/9),$$ i.e., $y=x/6+11/9$, and $$(61/50,303/250,1)times(1,12/10,0)=(-6/5,1,63/250).$$ Their intersection is therefore $$left(-frac16,1,-frac{11}9right)timesleft(-frac65,1,frac{63}{250}right) approx (1.474,1.509,1.033),$$ which we dehomogenize by dividing through by the last coordinate, yielding $Papprox(1.427,1.460)$. Intersecting the target lines, we have $$left(frac16,-1,frac{45}{60}right)timesleft(frac{12}{10},-1,0right) approx (0.750,0.900,1.033),$$ or $Qapprox (0.726,0.871)$, from which $Delta x approx -0.70$, $Delta yapprox -0.59$. The shifted parabola is therefore (to two decimal places) $$y = f(x+0.70)-0.59 = (x+0.70)-5(0.5(x+0.70)-0.65)^2-0.59.$$
answered Nov 24 '18 at 0:55
amdamd
29.4k21050
edited Nov 24 '18 at 3:13
answered Nov 24 '18 at 0:55
amdamd
29.4k21050
answered Nov 24 '18 at 0:55
amdamd
29.4k21050
answered Nov 24 '18 at 0:55
amdamd
29.4k21050
29.4k21050
Hello great thanks for answer. But I have impression like now after your help I know little bit less. Is it really f’(x)=1 - 5(0.5x + 0.65). I was sure it should be f’(x)=1 - 2*5(0.5x + 0.65). But now I am not sure because your solution also works. So I am in great big black hole :) What’s wrong?
– pajczur
Nov 24 '18 at 3:11
@pajczur You forgot to multiply by $0.5$ for the derivative of the parenthesized expression. You might want to review the chain rule.
– amd
Nov 24 '18 at 3:12
Yes chain rule would be great. What should I multiply by $0.5$ ? Exactly which part?
– pajczur
Nov 24 '18 at 3:17
Ok sorry my fault, I’ve found it thanks
– pajczur
Nov 24 '18 at 4:02
add a comment |
Hello great thanks for answer. But I have impression like now after your help I know little bit less. Is it really f’(x)=1 - 5(0.5x + 0.65). I was sure it should be f’(x)=1 - 2*5(0.5x + 0.65). But now I am not sure because your solution also works. So I am in great big black hole :) What’s wrong?
– pajczur
Nov 24 '18 at 3:11
@pajczur You forgot to multiply by $0.5$ for the derivative of the parenthesized expression. You might want to review the chain rule.
– amd
Nov 24 '18 at 3:12
Yes chain rule would be great. What should I multiply by $0.5$ ? Exactly which part?
– pajczur
Nov 24 '18 at 3:17
Ok sorry my fault, I’ve found it thanks
– pajczur
Nov 24 '18 at 4:02
Hello great thanks for answer. But I have impression like now after your help I know little bit less. Is it really f’(x)=1 - 5(0.5x + 0.65). I was sure it should be f’(x)=1 - 2*5(0.5x + 0.65). But now I am not sure because your solution also works. So I am in great big black hole :) What’s wrong?
– pajczur
Nov 24 '18 at 3:11
Hello great thanks for answer. But I have impression like now after your help I know little bit less. Is it really f’(x)=1 - 5(0.5x + 0.65). I was sure it should be f’(x)=1 - 2*5(0.5x + 0.65). But now I am not sure because your solution also works. So I am in great big black hole :) What’s wrong?
– pajczur
Nov 24 '18 at 3:11
@pajczur You forgot to multiply by $0.5$ for the derivative of the parenthesized expression. You might want to review the chain rule.
– amd
Nov 24 '18 at 3:12
@pajczur You forgot to multiply by $0.5$ for the derivative of the parenthesized expression. You might want to review the chain rule.
– amd
Nov 24 '18 at 3:12
Yes chain rule would be great. What should I multiply by $0.5$ ? Exactly which part?
– pajczur
Nov 24 '18 at 3:17
Yes chain rule would be great. What should I multiply by $0.5$ ? Exactly which part?
– pajczur
Nov 24 '18 at 3:17
Ok sorry my fault, I’ve found it thanks
– pajczur
Nov 24 '18 at 4:02
Ok sorry my fault, I’ve found it thanks
– pajczur
Nov 24 '18 at 4:02
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008672%2fmove-parabola-to-make-it-tangent-to-lines%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
