Using the determinants delta and Delta , determine the…
Using the determinants $\delta$ and $\Delta$, determine the geometric meaning of: $1^\circ$ $x^2-4xy+3y^2-8x+14y+15=0$; $2^\circ$ $2x^2+3xy-5y^2=0$; $3^\circ$ $x^2-xy-y^2-x-y=0$.
Explanation of Method
The determinants δ and Δ identify two line pairs and a hyperbola.
Step 1: Invariants
\[ \delta=\begin{vmatrix}A&\tfrac B2\\\tfrac B2&C\end{vmatrix},\qquad \Delta=\begin{vmatrix}A&\tfrac B2&\tfrac D2\\\tfrac B2&C&\tfrac E2\\\tfrac D2&\tfrac E2&F\end{vmatrix} \]
| $\Delta\ne0$ | $\Delta=0$ | |
|---|---|---|
| $\delta\gt 0$ | ellipse (real if $A\Delta\lt 0$, imaginary if $A\Delta\gt 0$) | a single point |
| $\delta\lt 0$ | hyperbola | two intersecting lines |
| $\delta=0$ | parabola | two parallel lines (real, coincident or imaginary) |
Step 2: $1^\circ$
\[ \delta=3-4=-1,\qquad \Delta=\begin{vmatrix}1&-2&-4\\-2&3&7\\-4&7&15\end{vmatrix}=-4-4+8=0 . \]
So there are two intersecting lines. Matching coefficients: $(x-y-3)(x-3y-5)=0,\qquad\text{intersection }(2,-1)$.
Step 3: $2^\circ$
Here $\delta=-10-\tfrac94=-\tfrac{49}4$ and $\Delta=0$, since the last row is zero. The roots $x=\tfrac{-3y\pm7y}{4}$ give
\[(x-y)(2x+5y)=0:\quad y=x\ \text{ and }\ y=-\tfrac25x\]
Step 4: $3^\circ$
Here $\delta=-1-\tfrac14=-\tfrac54$ and $\Delta=-\tfrac14\ne0$, so the curve is a hyperbola. Its centre is $\left(\tfrac15,-\tfrac35\right)$ and $F'=\tfrac12\left(-\tfrac15+\tfrac35\right)=\tfrac15$. The eigenvalues are $\pm\tfrac{\sqrt5}2$, because $A+C=0$:
\[ \tfrac{\sqrt5}2X^2-\tfrac{\sqrt5}2Y^2=-\tfrac15\ \Longrightarrow\ \frac{Y^2}{a^2}-\frac{X^2}{a^2}=1,\quad a^2=\frac{2}{5\sqrt5}=\frac{2\sqrt5}{25}. \]
\[\text{equilateral hyperbola }(a=b),\quad e=\sqrt2\]
Result
\[\boxed{\,(x-y-3)(x-3y-5)=0,\qquad\text{intersection }(2,-1)\,}\]
\[\boxed{\,(x-y)(2x+5y)=0:\quad y=x\ \text{ and }\ y=-\tfrac25x\,}\]
\[\boxed{\,\text{equilateral hyperbola }(a=b),\quad e=\sqrt2\,}\]
Remark
$A+C=0$ is exactly the condition for an equilateral (rectangular) hyperbola, since then the eigenvalues are opposite.