Finding equation of a curve that a parametric equation of a line is tangent to












0












$begingroup$


I have the line $y=xcdotfrac{-a}{-3a^2+4}$.
I want to find the curve that this line is tangent to.
More info: the tangency point needs to be at coordinates $(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16})$



Clarifications:



The line should always be tangent to the curve, even when you plug in different values in the parameter.



Also, the curve must be constant, meaning that it should not be expressed via parameters.






parametric







share|cite|improve this question











$endgroup$












  • $begingroup$
    What sort of curve? A circle, a parabola or what?
    $endgroup$
    – Michael Hoppe
    Dec 10 '18 at 19:48










  • $begingroup$
    I think there's only one curve that my info applies to, but it indeed should be a function.
    $endgroup$
    – TreasureGhost
    Dec 10 '18 at 19:56






  • 1




    $begingroup$
    @TreasureGhost Not true. Just draw a perpendicular to the line through the tangent point. Choose any point on that line and draw the circle with the center in the chosen point, and the radius equal to the distance from the chosen point to the tangent point. This circle is tangent to the line. Or choose $a=0$. Then the tangent point is $(0,0)$ and the tangent line is $y=0$. I can choose $f(x)=x^2$, or $3x^2$, or $x^5+pi x^3$, or $cos(x)-1$, and so on
    $endgroup$
    – Andrei
    Dec 10 '18 at 20:01










  • $begingroup$
    Is it always tangent to the line as the parameter changes values, though? I meant that the line should be always tangent to the curve. Similarly: $2ax-a^2$ is always tangent to $x^2$ as the parameter changes values. I'll add some clarifications.
    $endgroup$
    – TreasureGhost
    Dec 10 '18 at 20:10








  • 1




    $begingroup$
    You know the slope of the tangent at every point on the curve. Does this suggest a differential equation that you can try to solve?
    $endgroup$
    – amd
    Dec 10 '18 at 23:24
















0












$begingroup$


I have the line $y=xcdotfrac{-a}{-3a^2+4}$.
I want to find the curve that this line is tangent to.
More info: the tangency point needs to be at coordinates $(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16})$



Clarifications:



The line should always be tangent to the curve, even when you plug in different values in the parameter.



Also, the curve must be constant, meaning that it should not be expressed via parameters.






parametric







share|cite|improve this question











$endgroup$












  • $begingroup$
    What sort of curve? A circle, a parabola or what?
    $endgroup$
    – Michael Hoppe
    Dec 10 '18 at 19:48










  • $begingroup$
    I think there's only one curve that my info applies to, but it indeed should be a function.
    $endgroup$
    – TreasureGhost
    Dec 10 '18 at 19:56






  • 1




    $begingroup$
    @TreasureGhost Not true. Just draw a perpendicular to the line through the tangent point. Choose any point on that line and draw the circle with the center in the chosen point, and the radius equal to the distance from the chosen point to the tangent point. This circle is tangent to the line. Or choose $a=0$. Then the tangent point is $(0,0)$ and the tangent line is $y=0$. I can choose $f(x)=x^2$, or $3x^2$, or $x^5+pi x^3$, or $cos(x)-1$, and so on
    $endgroup$
    – Andrei
    Dec 10 '18 at 20:01










  • $begingroup$
    Is it always tangent to the line as the parameter changes values, though? I meant that the line should be always tangent to the curve. Similarly: $2ax-a^2$ is always tangent to $x^2$ as the parameter changes values. I'll add some clarifications.
    $endgroup$
    – TreasureGhost
    Dec 10 '18 at 20:10








  • 1




    $begingroup$
    You know the slope of the tangent at every point on the curve. Does this suggest a differential equation that you can try to solve?
    $endgroup$
    – amd
    Dec 10 '18 at 23:24














0












0








0





$begingroup$


I have the line $y=xcdotfrac{-a}{-3a^2+4}$.
I want to find the curve that this line is tangent to.
More info: the tangency point needs to be at coordinates $(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16})$



Clarifications:



The line should always be tangent to the curve, even when you plug in different values in the parameter.



Also, the curve must be constant, meaning that it should not be expressed via parameters.






parametric







share|cite|improve this question











$endgroup$




I have the line $y=xcdotfrac{-a}{-3a^2+4}$.
I want to find the curve that this line is tangent to.
More info: the tangency point needs to be at coordinates $(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16})$



Clarifications:



The line should always be tangent to the curve, even when you plug in different values in the parameter.



Also, the curve must be constant, meaning that it should not be expressed via parameters.






parametric




parametric






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 20:12







TreasureGhost

















asked Dec 10 '18 at 19:46









TreasureGhostTreasureGhost

152




152












  • $begingroup$
    What sort of curve? A circle, a parabola or what?
    $endgroup$
    – Michael Hoppe
    Dec 10 '18 at 19:48










  • $begingroup$
    I think there's only one curve that my info applies to, but it indeed should be a function.
    $endgroup$
    – TreasureGhost
    Dec 10 '18 at 19:56






  • 1




    $begingroup$
    @TreasureGhost Not true. Just draw a perpendicular to the line through the tangent point. Choose any point on that line and draw the circle with the center in the chosen point, and the radius equal to the distance from the chosen point to the tangent point. This circle is tangent to the line. Or choose $a=0$. Then the tangent point is $(0,0)$ and the tangent line is $y=0$. I can choose $f(x)=x^2$, or $3x^2$, or $x^5+pi x^3$, or $cos(x)-1$, and so on
    $endgroup$
    – Andrei
    Dec 10 '18 at 20:01










  • $begingroup$
    Is it always tangent to the line as the parameter changes values, though? I meant that the line should be always tangent to the curve. Similarly: $2ax-a^2$ is always tangent to $x^2$ as the parameter changes values. I'll add some clarifications.
    $endgroup$
    – TreasureGhost
    Dec 10 '18 at 20:10








  • 1




    $begingroup$
    You know the slope of the tangent at every point on the curve. Does this suggest a differential equation that you can try to solve?
    $endgroup$
    – amd
    Dec 10 '18 at 23:24


















  • $begingroup$
    What sort of curve? A circle, a parabola or what?
    $endgroup$
    – Michael Hoppe
    Dec 10 '18 at 19:48










  • $begingroup$
    I think there's only one curve that my info applies to, but it indeed should be a function.
    $endgroup$
    – TreasureGhost
    Dec 10 '18 at 19:56






  • 1




    $begingroup$
    @TreasureGhost Not true. Just draw a perpendicular to the line through the tangent point. Choose any point on that line and draw the circle with the center in the chosen point, and the radius equal to the distance from the chosen point to the tangent point. This circle is tangent to the line. Or choose $a=0$. Then the tangent point is $(0,0)$ and the tangent line is $y=0$. I can choose $f(x)=x^2$, or $3x^2$, or $x^5+pi x^3$, or $cos(x)-1$, and so on
    $endgroup$
    – Andrei
    Dec 10 '18 at 20:01










  • $begingroup$
    Is it always tangent to the line as the parameter changes values, though? I meant that the line should be always tangent to the curve. Similarly: $2ax-a^2$ is always tangent to $x^2$ as the parameter changes values. I'll add some clarifications.
    $endgroup$
    – TreasureGhost
    Dec 10 '18 at 20:10








  • 1




    $begingroup$
    You know the slope of the tangent at every point on the curve. Does this suggest a differential equation that you can try to solve?
    $endgroup$
    – amd
    Dec 10 '18 at 23:24
















$begingroup$
What sort of curve? A circle, a parabola or what?
$endgroup$
– Michael Hoppe
Dec 10 '18 at 19:48




$begingroup$
What sort of curve? A circle, a parabola or what?
$endgroup$
– Michael Hoppe
Dec 10 '18 at 19:48












$begingroup$
I think there's only one curve that my info applies to, but it indeed should be a function.
$endgroup$
– TreasureGhost
Dec 10 '18 at 19:56




$begingroup$
I think there's only one curve that my info applies to, but it indeed should be a function.
$endgroup$
– TreasureGhost
Dec 10 '18 at 19:56




1




1




$begingroup$
@TreasureGhost Not true. Just draw a perpendicular to the line through the tangent point. Choose any point on that line and draw the circle with the center in the chosen point, and the radius equal to the distance from the chosen point to the tangent point. This circle is tangent to the line. Or choose $a=0$. Then the tangent point is $(0,0)$ and the tangent line is $y=0$. I can choose $f(x)=x^2$, or $3x^2$, or $x^5+pi x^3$, or $cos(x)-1$, and so on
$endgroup$
– Andrei
Dec 10 '18 at 20:01




$begingroup$
@TreasureGhost Not true. Just draw a perpendicular to the line through the tangent point. Choose any point on that line and draw the circle with the center in the chosen point, and the radius equal to the distance from the chosen point to the tangent point. This circle is tangent to the line. Or choose $a=0$. Then the tangent point is $(0,0)$ and the tangent line is $y=0$. I can choose $f(x)=x^2$, or $3x^2$, or $x^5+pi x^3$, or $cos(x)-1$, and so on
$endgroup$
– Andrei
Dec 10 '18 at 20:01












$begingroup$
Is it always tangent to the line as the parameter changes values, though? I meant that the line should be always tangent to the curve. Similarly: $2ax-a^2$ is always tangent to $x^2$ as the parameter changes values. I'll add some clarifications.
$endgroup$
– TreasureGhost
Dec 10 '18 at 20:10






$begingroup$
Is it always tangent to the line as the parameter changes values, though? I meant that the line should be always tangent to the curve. Similarly: $2ax-a^2$ is always tangent to $x^2$ as the parameter changes values. I'll add some clarifications.
$endgroup$
– TreasureGhost
Dec 10 '18 at 20:10






1




1




$begingroup$
You know the slope of the tangent at every point on the curve. Does this suggest a differential equation that you can try to solve?
$endgroup$
– amd
Dec 10 '18 at 23:24




$begingroup$
You know the slope of the tangent at every point on the curve. Does this suggest a differential equation that you can try to solve?
$endgroup$
– amd
Dec 10 '18 at 23:24










1 Answer
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the tangency point needs to be at coordinates $(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16})$




The point of tangency is a point on your curve. So for each value of $a$, the point $$left(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16}right)$$



is a point on your curve. I.e., a parametric equation for your curve is $$(x(t), y(t) ) =left(frac{tleft(-3t^2+4right)}{-9t^2+16},-frac{t^2}{-9t^2+16}right)$$



If this curve doesn't meet your tangency condition (you can check that yourself), then you are out-of-luck. No curve can satisfy your conditions.






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    $begingroup$


    the tangency point needs to be at coordinates $(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16})$




    The point of tangency is a point on your curve. So for each value of $a$, the point $$left(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16}right)$$



    is a point on your curve. I.e., a parametric equation for your curve is $$(x(t), y(t) ) =left(frac{tleft(-3t^2+4right)}{-9t^2+16},-frac{t^2}{-9t^2+16}right)$$



    If this curve doesn't meet your tangency condition (you can check that yourself), then you are out-of-luck. No curve can satisfy your conditions.






    share|cite|improve this answer









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      0












      $begingroup$


      the tangency point needs to be at coordinates $(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16})$




      The point of tangency is a point on your curve. So for each value of $a$, the point $$left(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16}right)$$



      is a point on your curve. I.e., a parametric equation for your curve is $$(x(t), y(t) ) =left(frac{tleft(-3t^2+4right)}{-9t^2+16},-frac{t^2}{-9t^2+16}right)$$



      If this curve doesn't meet your tangency condition (you can check that yourself), then you are out-of-luck. No curve can satisfy your conditions.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$


        the tangency point needs to be at coordinates $(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16})$




        The point of tangency is a point on your curve. So for each value of $a$, the point $$left(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16}right)$$



        is a point on your curve. I.e., a parametric equation for your curve is $$(x(t), y(t) ) =left(frac{tleft(-3t^2+4right)}{-9t^2+16},-frac{t^2}{-9t^2+16}right)$$



        If this curve doesn't meet your tangency condition (you can check that yourself), then you are out-of-luck. No curve can satisfy your conditions.






        share|cite|improve this answer









        $endgroup$




        the tangency point needs to be at coordinates $(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16})$




        The point of tangency is a point on your curve. So for each value of $a$, the point $$left(frac{aleft(-3a^2+4right)}{-9a^2+16},-frac{a^2}{-9a^2+16}right)$$



        is a point on your curve. I.e., a parametric equation for your curve is $$(x(t), y(t) ) =left(frac{tleft(-3t^2+4right)}{-9t^2+16},-frac{t^2}{-9t^2+16}right)$$



        If this curve doesn't meet your tangency condition (you can check that yourself), then you are out-of-luck. No curve can satisfy your conditions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 2:54









        Paul SinclairPaul Sinclair

        20.7k21543




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