Proving y^2=-4x

May 2010
14
0
I am given equations with coefficients related by a common ratio, thus a geometric sequence, such as

2x+4y=8
3x+12y=48

I plotted numerous lines which obey this geometric pattern and discovered that there is a curve which through trial and error I discovered to be y=(-4x)^(1/2). I also discovered that each line was tangential to the curve.

I proceeded to create a general proof of the solutions of equations obeying this general form:

ax+ary=ar^2
bx+bdy=bd^2

I discovered the solution of x and y using simultaneous equations (I am not posting the answer here because this is a portfolio task and direct answers are not allowed).

I am now stuck on proving the curve; can anybody tell me if my approach above is correct in working the relations and to go where from this point.
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, chaosier!

I am given equations with coefficients related by a common ratio,
thus a geometric sequence, such as: .\(\displaystyle \begin{array}{ccc}2x+4y&=&8 \\ 3x+12y&=&48 \end{array}\)

I plotted numerous lines which obey this geometric pattern
. . and discovered that there is a curve which,
. . through trial and error, I found to be: .\(\displaystyle y\:=\:(-4x)^{\frac{1}{2}}\) . . . . I agree!
I also found that each line was tangential to the curve.

I proceeded to create a general proof of the solutions
. . of equations with this general form: .\(\displaystyle \begin{array}{cccc}ax+apy &=&ap^2 & [1] \\ bx+bqy &=& bq^2 & [2] \end{array}\)

I found the solution of \(\displaystyle x\) and \(\displaystyle y\) using simultaneous equations.

I am now stuck on proving the curve.
Can anybody tell me if my approach above is correct
. . and where to go from this point?

\(\displaystyle \begin{array}{ccccccc}\text{Divide [1] by }a\!: & x + py &=& p^2 & [3] \\ \text{Divide [2] by }b\!: & x + qy &=& q^2 & [4]\end{array}\)

Subtract [3] - [4]: .\(\displaystyle py - qy \:=\:p^2-q^2 \quad\Rightarrow\quad (p-q)y \:=\:(p-q)(p+q) \)

. . Hence: .\(\displaystyle y \:=\:p+q\)

Substitute into [3]: .\(\displaystyle x + p(p+q) \:=\:p^2 \quad\Rightarrow\quad x \:=\:pq\)


We have parametric equations for the intersection
. . of two members of this family of curves.

. . . . . \(\displaystyle \begin{Bmatrix} x &=& -pq \\ y &=& p+q \end{Bmatrix}\)


To determine the "envelope" of this family of curves, let \(\displaystyle q \to p.\)

. . Hence, we have: .\(\displaystyle \begin{array}{ccccc}x &=& -p^2 & [5]\\ y &=& 2p & [6]\end{array}\)


Eliminate the parameter:
. . From [6]: .\(\displaystyle y \:=\:2p \quad\Rightarrow\quad p \:=\:\tfrac{y}{2}\)
. . Substitute into [5]: .\(\displaystyle x \:=\:-\left(\tfrac{y}{2}\right)^2 \quad\Rightarrow\quad \boxed{y^2 \:=\:-4x}\)

 
  • Like
Reactions: chaosier
May 2010
14
0
Soroban could you please elaborate a bit on q->p?

Is it because q and p only represent two ratios when there could be any ratio from q to p, and from letting q tend to p you are proving that the graphical pattern that manifests from these results are true for all ratios and by proving this generate a general form for the graphical pattern.