Maximum number of common normal of $y^2=4ax$ and $x^2 = 4by$












2












$begingroup$



Maximum number of common normal of $y^2=4ax$ and $x^2 = 4by$




Attempt: Equation of normal to the curve $y^2=4ax$ in slope form $y=mx-2am-am^3.$



Now above equation is also normal to $x^2=4by$



I could not understand how to solve further, thanks.






analytic-geometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    eq of normal for parabola of form $x^2=4by$ is $y=mx+2b + frac{b}{m^2}$.Compare the 2 equations
    $endgroup$
    – Abhash Jha
    May 10 '17 at 14:43
















2












$begingroup$



Maximum number of common normal of $y^2=4ax$ and $x^2 = 4by$




Attempt: Equation of normal to the curve $y^2=4ax$ in slope form $y=mx-2am-am^3.$



Now above equation is also normal to $x^2=4by$



I could not understand how to solve further, thanks.






analytic-geometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    eq of normal for parabola of form $x^2=4by$ is $y=mx+2b + frac{b}{m^2}$.Compare the 2 equations
    $endgroup$
    – Abhash Jha
    May 10 '17 at 14:43














2












2








2


2



$begingroup$



Maximum number of common normal of $y^2=4ax$ and $x^2 = 4by$




Attempt: Equation of normal to the curve $y^2=4ax$ in slope form $y=mx-2am-am^3.$



Now above equation is also normal to $x^2=4by$



I could not understand how to solve further, thanks.






analytic-geometry







share|cite|improve this question











$endgroup$





Maximum number of common normal of $y^2=4ax$ and $x^2 = 4by$




Attempt: Equation of normal to the curve $y^2=4ax$ in slope form $y=mx-2am-am^3.$



Now above equation is also normal to $x^2=4by$



I could not understand how to solve further, thanks.






analytic-geometry




analytic-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 10:48









amWhy

1




1










asked May 10 '17 at 14:25









DXTDXT

5,5722630




5,5722630








  • 1




    $begingroup$
    eq of normal for parabola of form $x^2=4by$ is $y=mx+2b + frac{b}{m^2}$.Compare the 2 equations
    $endgroup$
    – Abhash Jha
    May 10 '17 at 14:43














  • 1




    $begingroup$
    eq of normal for parabola of form $x^2=4by$ is $y=mx+2b + frac{b}{m^2}$.Compare the 2 equations
    $endgroup$
    – Abhash Jha
    May 10 '17 at 14:43








1




1




$begingroup$
eq of normal for parabola of form $x^2=4by$ is $y=mx+2b + frac{b}{m^2}$.Compare the 2 equations
$endgroup$
– Abhash Jha
May 10 '17 at 14:43




$begingroup$
eq of normal for parabola of form $x^2=4by$ is $y=mx+2b + frac{b}{m^2}$.Compare the 2 equations
$endgroup$
– Abhash Jha
May 10 '17 at 14:43










1 Answer
1






active

oldest

votes


















2












$begingroup$

In general, a normal to $y^2=4ax$ has the form



$$tx+y-2at-at^3=0$$



and a normal to $x^2=4by$ has the form



$$x+sy-2bs-bs^3=0$$



A common normal of the two parabolae has its equation expressible in both the above two forms. Obviously, $sne0$ and $tne0$.



The second equation can be written as



$$tx+sty-2bst-bs^3t=0$$



Therefore, $st=1$ and



begin{align*}
-2at-at^3&=-2bst-bs^3t\
&=-2b-bs^2\
&=-2b-frac{b}{t^2}\
2at^3+at^5&=2bt^2+b
end{align*}



Let $f(t)=at^5+2at^3-2bt^2-b$. Then $f(-t)=-at^5-2at^3-2bt^2-b$.



If $a$ and $b$ are of the same sign, then the coefficients of $f(t)$ has one sign change and the coefficients of $f(-t)$ has no sign change. So $f(t)$ has one positive root and no negative root.



If $a$ and $b$ are of opposite signs, then the coefficients of $f(t)$ has no sign change and the coefficients of $f(-t)$ has one sign change. So $f(t)$ has no positive root and one negative root.



Therefore, $f(t)$ has only one real root. The two parabolae have at most one common normal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What is the meaning of "one sign change"? Please explain.
    $endgroup$
    – jayant98
    Dec 17 '18 at 10:21










  • $begingroup$
    From $2a$ to $2b$, there is one sign change.
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:24










  • $begingroup$
    You mean to from $2at^3$ to $2bt^2$?
    $endgroup$
    – jayant98
    Dec 17 '18 at 10:35












  • $begingroup$
    Yes. For the case of same sign
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:38






  • 1




    $begingroup$
    en.wikipedia.org/wiki/Descartes%27_rule_of_signs
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:52











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

In general, a normal to $y^2=4ax$ has the form



$$tx+y-2at-at^3=0$$



and a normal to $x^2=4by$ has the form



$$x+sy-2bs-bs^3=0$$



A common normal of the two parabolae has its equation expressible in both the above two forms. Obviously, $sne0$ and $tne0$.



The second equation can be written as



$$tx+sty-2bst-bs^3t=0$$



Therefore, $st=1$ and



begin{align*}
-2at-at^3&=-2bst-bs^3t\
&=-2b-bs^2\
&=-2b-frac{b}{t^2}\
2at^3+at^5&=2bt^2+b
end{align*}



Let $f(t)=at^5+2at^3-2bt^2-b$. Then $f(-t)=-at^5-2at^3-2bt^2-b$.



If $a$ and $b$ are of the same sign, then the coefficients of $f(t)$ has one sign change and the coefficients of $f(-t)$ has no sign change. So $f(t)$ has one positive root and no negative root.



If $a$ and $b$ are of opposite signs, then the coefficients of $f(t)$ has no sign change and the coefficients of $f(-t)$ has one sign change. So $f(t)$ has no positive root and one negative root.



Therefore, $f(t)$ has only one real root. The two parabolae have at most one common normal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What is the meaning of "one sign change"? Please explain.
    $endgroup$
    – jayant98
    Dec 17 '18 at 10:21










  • $begingroup$
    From $2a$ to $2b$, there is one sign change.
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:24










  • $begingroup$
    You mean to from $2at^3$ to $2bt^2$?
    $endgroup$
    – jayant98
    Dec 17 '18 at 10:35












  • $begingroup$
    Yes. For the case of same sign
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:38






  • 1




    $begingroup$
    en.wikipedia.org/wiki/Descartes%27_rule_of_signs
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:52
















2












$begingroup$

In general, a normal to $y^2=4ax$ has the form



$$tx+y-2at-at^3=0$$



and a normal to $x^2=4by$ has the form



$$x+sy-2bs-bs^3=0$$



A common normal of the two parabolae has its equation expressible in both the above two forms. Obviously, $sne0$ and $tne0$.



The second equation can be written as



$$tx+sty-2bst-bs^3t=0$$



Therefore, $st=1$ and



begin{align*}
-2at-at^3&=-2bst-bs^3t\
&=-2b-bs^2\
&=-2b-frac{b}{t^2}\
2at^3+at^5&=2bt^2+b
end{align*}



Let $f(t)=at^5+2at^3-2bt^2-b$. Then $f(-t)=-at^5-2at^3-2bt^2-b$.



If $a$ and $b$ are of the same sign, then the coefficients of $f(t)$ has one sign change and the coefficients of $f(-t)$ has no sign change. So $f(t)$ has one positive root and no negative root.



If $a$ and $b$ are of opposite signs, then the coefficients of $f(t)$ has no sign change and the coefficients of $f(-t)$ has one sign change. So $f(t)$ has no positive root and one negative root.



Therefore, $f(t)$ has only one real root. The two parabolae have at most one common normal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What is the meaning of "one sign change"? Please explain.
    $endgroup$
    – jayant98
    Dec 17 '18 at 10:21










  • $begingroup$
    From $2a$ to $2b$, there is one sign change.
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:24










  • $begingroup$
    You mean to from $2at^3$ to $2bt^2$?
    $endgroup$
    – jayant98
    Dec 17 '18 at 10:35












  • $begingroup$
    Yes. For the case of same sign
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:38






  • 1




    $begingroup$
    en.wikipedia.org/wiki/Descartes%27_rule_of_signs
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:52














2












2








2





$begingroup$

In general, a normal to $y^2=4ax$ has the form



$$tx+y-2at-at^3=0$$



and a normal to $x^2=4by$ has the form



$$x+sy-2bs-bs^3=0$$



A common normal of the two parabolae has its equation expressible in both the above two forms. Obviously, $sne0$ and $tne0$.



The second equation can be written as



$$tx+sty-2bst-bs^3t=0$$



Therefore, $st=1$ and



begin{align*}
-2at-at^3&=-2bst-bs^3t\
&=-2b-bs^2\
&=-2b-frac{b}{t^2}\
2at^3+at^5&=2bt^2+b
end{align*}



Let $f(t)=at^5+2at^3-2bt^2-b$. Then $f(-t)=-at^5-2at^3-2bt^2-b$.



If $a$ and $b$ are of the same sign, then the coefficients of $f(t)$ has one sign change and the coefficients of $f(-t)$ has no sign change. So $f(t)$ has one positive root and no negative root.



If $a$ and $b$ are of opposite signs, then the coefficients of $f(t)$ has no sign change and the coefficients of $f(-t)$ has one sign change. So $f(t)$ has no positive root and one negative root.



Therefore, $f(t)$ has only one real root. The two parabolae have at most one common normal.






share|cite|improve this answer











$endgroup$



In general, a normal to $y^2=4ax$ has the form



$$tx+y-2at-at^3=0$$



and a normal to $x^2=4by$ has the form



$$x+sy-2bs-bs^3=0$$



A common normal of the two parabolae has its equation expressible in both the above two forms. Obviously, $sne0$ and $tne0$.



The second equation can be written as



$$tx+sty-2bst-bs^3t=0$$



Therefore, $st=1$ and



begin{align*}
-2at-at^3&=-2bst-bs^3t\
&=-2b-bs^2\
&=-2b-frac{b}{t^2}\
2at^3+at^5&=2bt^2+b
end{align*}



Let $f(t)=at^5+2at^3-2bt^2-b$. Then $f(-t)=-at^5-2at^3-2bt^2-b$.



If $a$ and $b$ are of the same sign, then the coefficients of $f(t)$ has one sign change and the coefficients of $f(-t)$ has no sign change. So $f(t)$ has one positive root and no negative root.



If $a$ and $b$ are of opposite signs, then the coefficients of $f(t)$ has no sign change and the coefficients of $f(-t)$ has one sign change. So $f(t)$ has no positive root and one negative root.



Therefore, $f(t)$ has only one real root. The two parabolae have at most one common normal.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 16 '18 at 17:55









Ng Chung Tak

14.4k31334




14.4k31334










answered May 10 '17 at 15:45









CY AriesCY Aries

16.7k11743




16.7k11743












  • $begingroup$
    What is the meaning of "one sign change"? Please explain.
    $endgroup$
    – jayant98
    Dec 17 '18 at 10:21










  • $begingroup$
    From $2a$ to $2b$, there is one sign change.
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:24










  • $begingroup$
    You mean to from $2at^3$ to $2bt^2$?
    $endgroup$
    – jayant98
    Dec 17 '18 at 10:35












  • $begingroup$
    Yes. For the case of same sign
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:38






  • 1




    $begingroup$
    en.wikipedia.org/wiki/Descartes%27_rule_of_signs
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:52


















  • $begingroup$
    What is the meaning of "one sign change"? Please explain.
    $endgroup$
    – jayant98
    Dec 17 '18 at 10:21










  • $begingroup$
    From $2a$ to $2b$, there is one sign change.
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:24










  • $begingroup$
    You mean to from $2at^3$ to $2bt^2$?
    $endgroup$
    – jayant98
    Dec 17 '18 at 10:35












  • $begingroup$
    Yes. For the case of same sign
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:38






  • 1




    $begingroup$
    en.wikipedia.org/wiki/Descartes%27_rule_of_signs
    $endgroup$
    – CY Aries
    Dec 17 '18 at 10:52
















$begingroup$
What is the meaning of "one sign change"? Please explain.
$endgroup$
– jayant98
Dec 17 '18 at 10:21




$begingroup$
What is the meaning of "one sign change"? Please explain.
$endgroup$
– jayant98
Dec 17 '18 at 10:21












$begingroup$
From $2a$ to $2b$, there is one sign change.
$endgroup$
– CY Aries
Dec 17 '18 at 10:24




$begingroup$
From $2a$ to $2b$, there is one sign change.
$endgroup$
– CY Aries
Dec 17 '18 at 10:24












$begingroup$
You mean to from $2at^3$ to $2bt^2$?
$endgroup$
– jayant98
Dec 17 '18 at 10:35






$begingroup$
You mean to from $2at^3$ to $2bt^2$?
$endgroup$
– jayant98
Dec 17 '18 at 10:35














$begingroup$
Yes. For the case of same sign
$endgroup$
– CY Aries
Dec 17 '18 at 10:38




$begingroup$
Yes. For the case of same sign
$endgroup$
– CY Aries
Dec 17 '18 at 10:38




1




1




$begingroup$
en.wikipedia.org/wiki/Descartes%27_rule_of_signs
$endgroup$
– CY Aries
Dec 17 '18 at 10:52




$begingroup$
en.wikipedia.org/wiki/Descartes%27_rule_of_signs
$endgroup$
– CY Aries
Dec 17 '18 at 10:52


















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