How to prove the pair of second degree equations enclose a square?

Question

Show that the two pairs of second degree equations $12x^2+7xy-12y^2=0$ and $12x^2+7xy-12y^2-x+7y-1=0$ form a square.

I tried to do this by calculating the slope and multiplying to give $-1$ and show perpendicular lines but this is difficult as the equation is a second degree equation.

The graph looks like this-

So,how do I approach the problem?

What is the general method for solving this type of problems?

Thanks for any help!!

Answer 1

For example:

$$12x^2+7xy-12y^2=0\;\;\;\text{is a quadratic in $x$ whose solutions are:}$$

$$x_{1,2}=\frac{-7y\pm\sqrt{49y^2+576y^2}}{24}=\frac{-7y\pm25y}{24}=\begin{cases}-\cfrac{32y}{24}=-\cfrac43y\\{}\\\cfrac{18y}{24}=\cfrac34y\end{cases}$$

and the solution lines to the above quadratic are $\;x=-\frac43y\;,\;\;x=\frac34y\;$ . Observe we already have two perpendicular lines. Can you continue?

Answer 2

• First equation: $$(4x-3y)(3x+4y)=0$$ which is a pair of perpendicular lines.

• Second equation: $$(4x-3y+a)(3x+4y+b)=0$$ which is a pair of perpendicular lines with $$ \left \{ \begin{array}{rcl} 3a+4b &= -1 \\ 4a-3b &= 7 \\ ab &= -1 \end{array} \right. \implies (a,b)=(1,-1)$$

• $4x-3y=0$ and $4x-3y+1=0$ is parallel with distance of $\dfrac{1}{\sqrt{4^2+3^2}}=\dfrac{1}{5}$

• $3x+4y=0$ and $3x+4y-1=0$ is parallel with distance of $\dfrac{1}{\sqrt{3^2+4^2}}=\dfrac{1}{5}$

Hence, the four lines enclose a square of length $\dfrac{1}{5}$.