Matematické Fórum

stuart clark · 02. 02. 2012 04:55

The maximum no. of Common Normals of $y^2=4ax$ and $x^2=4by$ is equal to

options::

(i) $3$

(ii) $4$

(iii) $5$

(iv) $6$

vanok · 05. 02. 2012 18:43 — Editoval vanok (06. 02. 2012 08:23)

Hi ↑ stuart clark:,
The equation
$y^2=4ax$ , $(EQ1)$
give
$2yy'=4a$
$y'=\frac{2a}y$
So in the point $x = at^2$ , $y=2at$ of the parabola $(EQ1)$ we have
$y'=\frac 1t$
Thus a equation of the shape
$y=Ax+B$
is a tangent of our parabola,if
$A=\frac 1t$
and
$B= 2at -\frac 1t*at^2=at$
Finally $y=\frac xt +at$ is an equation of the tangent of the parabola $(EQ1)$ at the point $x = at^2$ , $y=2at$ ... for all t

The slope of the normal at the point $x = at^2$ , $y=2at$ , is $-t$ .
So an equation of the shape
$y=Cx+D$
is a normal of our parabola, if
$C=-t$
and
$D=2at-(-t)at^2=2at+at^3$
Finally
$y=-tx+2at+at^3$ , $(N1)$
is an equation of the normal of the parabola $(EQ1)$ for all t.

We notice that the equation $(EQ1)$ is "symmetric" with $x^2=4by$ , $(EQ2)$ .
What gives us mechanically an equation of the normal of the parabola $(EQ1)$ is
$x=-ty +2bt +bt^3$
and finally an equation of the normal of the parabola $(EQ2)$ for all t,
of the shape
$y=Ex+F$
is
$y=- \frac 1t x +2b +bt^2$ , $(N2)$ .

For $m= -t$ the equation $y=-tx+2at+at^3$ , $(N1)$ , we have
$y=mx-2am -am^3$ is a normal of parabola $(EQ1)$ for all m.

And $m= - \frac 1t$ the equation $y=- \frac 1t x +2b +bt^2$ , $(N2)$ , we have

$y=m x +2b +\frac b{m^2}$ , $(N2)$ is a normal of parabola $(EQ2)$ for all m.

Thus common normals verify:
$mx-2am -am^3=m x +2b +\frac b{m^2}$
what gives
$am^5+2am^3+2bm^2+b=0$ , an equation in m of the fifth degree
Conclusion: Option (iii):
The maximum no. of Common Normals of $y^2=4ax$ and $x^2=4by$ is equal to 5.

jardofpr

↑ vanok:

suppose

$\bigg(P(m)=A.\prod_{i=1}^{5}(m-m_{i}) \quad; i\neq j \Rightarrow m_{i} \neq m_{j}\,;\,i,j\in \{1,2,3,4,5\}\,;$
$m_{1}<m_{2}<m_{3}<m_{4}<m_{5}\,;\,m_{i} \in \mathbb{R}\, ,\, i=1,2,3,4,5 \,;\,A > 0\bigg)\Rightarrow$

$\Rightarrow \bigg( [P(m)<0 \,\, \forall m \in (-\infty, m_{1})\cup(m_{2},m_{3})\cup(m_{4},m_{5})]\wedge$ $\wedge[P(m)>0 \,\, \forall m \in (m_{1}, m_{2})\cup(m_{3},m_{4})\cup(m_{5},\infty)]\bigg) \stackrel{P\in C(\mathbb{R})}{\Rightarrow}$

$\stackrel{P\in C(\mathbb{R})}{\Rightarrow} \bigg(\exists c_{1},c_{2},c_{3},c_{4}\in\mathbb{R}\bigg):\bigg(\bigg[c_{i} \in (m_{i},m_{i+1})\bigg]\wedge\,\bigg[P'(c_{i})=0\,;i=1,2,3,4\bigg]\wedge$
$\wedge \bigg[P'(m)>0 \,\,\forall m \in (-\infty,c_{1})\cup(c_{2},c_{3})\cup (c_{4},\infty)\bigg]\wedge \bigg[P'(m)<0\,\,\forall m \in (c_{1},c_{2})\cup(c_{3},c_{4})\bigg]\bigg)\Rightarrow$

$\Rightarrow\bigg( P'(m)=A.\prod_{i=1}^{4}(m'-m'_{i})\,;\,i\neq j \Rightarrow m_{i} \neq m_{j}\,,\,m_{i}\in \mathbb{R}\,,\,i=1,2,3,4\,;\,A>0 \bigg)$

(the same line for $A<0$ , $>$ will turn to $<$ and $<$ to $>$ on the corresponding intervals of monotonicity of $P(x)$ , but the result is the same )

back to your final equation

$(am^5 + 2am^3 + 2bm^2+b)'= a\prod_{i=1}^{4}(m-m_{i})\,\,\,,\, m_{1},m_{2}\in \mathbb{R}\,,\,m_{3},m_{4} \in \mathbb{C} \,,\,\forall a,b \in \mathbb{R}-\{0\}$

nonzero roots

Autor skryl část textu. Zobrazit/skrýt!

for $b=0\vee a=0$ is easy to show that the number of common normals of the given equations is less then 5

$\bigg(V_{1}\Rightarrow V_{2}\Rightarrow \dots \Rightarrow V_{k}\Rightarrow 0\bigg) \Rightarrow \neg V_{1}$

all the options
$A = 0$
$\exists i \in \{1,2,3,4,5\}: m_{i} \in \mathbb{C}$
$\exists i,j \in \{1,2,3,4,5\}:i \neq j \wedge m_{i}=m_{j}$
imply that number of common normals of given equations is less than 5

is that correct?

vanok · 07. 02. 2012 10:27 — Editoval vanok (07. 02. 2012 12:28)

Hi ↑ jardofpr:,
The obtained answer is valid in $\mathbb{C}$ .

I left this question aside:
Which is the maximal number of real solutions of the equation obtained, for a, b suitable.

I have shall look at your approach, in this frame, in detail, to know if can have 1, 3 or 5 real solutions.

EDIT:Remark: I have think, that one can pose: $K=\frac ba$ ,(the cases a=0; b=0 are to be excluded) what gives:
$m^5+2m^3+2Km^2+K=0$ ,
And in first approach, the study of function
$f(m)=m^5+2m^3+2Km^2+K$ (in several cases particular) seems to me to confirm, that the equation finds admits only 1 real solution.

jardofpr · 07. 02. 2012 14:49

↑ vanok:

thanks, I was confused a bit,
because I couldn't even find more than one common normal for any ordered pair [a,b] :)
then i was wondering, if I should consider a,b,x,y alltogether as the variables, but still the problem was to find the fifth normal ..
i was not thinking about complex solutions, cause (I don't know why but) I supposed the task is to find the real solutions ...

stuart clark · 12. 02. 2012 17:33

Thanks vanok for nice explanation

and thanks jardofpr

Matematické Fórum

#1 02. 02. 2012 04:55

common normal

#2 05. 02. 2012 18:43 — Editoval vanok (06. 02. 2012 08:23)

Re: common normal

#3 07. 02. 2012 06:22 — Editoval jardofpr (07. 02. 2012 07:19)

Re: common normal

#4 07. 02. 2012 10:27 — Editoval vanok (07. 02. 2012 12:28)

Re: common normal

#5 07. 02. 2012 14:49

Re: common normal

#6 12. 02. 2012 17:33

Re: common normal

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