Proving y^2=-4x

**chaosier** · May 27th 2010, 04:48 AM

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** · May 27th 2010, 06:45 AM

Hello, chaosier!

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

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: . $y\:=\:(-4x)^{\frac{1}{2}}$ . . . . 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: . $\begin{array}{cccc}ax+apy &=&ap^2 & [1] \\ bx+bqy &=& bq^2 & [2] \end{array}$

I found the solution of $x$ and $y$ 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?

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

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

. . Hence: . $y \:=\:p+q$

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

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

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

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

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

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

**chaosier** · May 27th 2010, 06:52 AM

Soroban, you are a genius!

**chaosier** · May 27th 2010, 07:48 AM

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.

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