# Proving y^2=-4x

#### chaosier

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
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]: .$$\displaystyle py - qy \:=\^2-q^2 \quad\Rightarrow\quad (p-q)y \:=\p-q)(p+q)$$

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

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

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]: .$$\displaystyle x \:=\:-\left(\tfrac{y}{2}\right)^2 \quad\Rightarrow\quad \boxed{y^2 \:=\:-4x}$$

chaosier

#### chaosier

Soroban, you are a genius!

#### chaosier

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.