Chapter 8.3, Problem 71E

Solution

To Prove: A non-degenerate graph of the given equation is a hyperbola if $AC<0$ .

It has been shown that a non-degenerate graph of the given equation is a hyperbola if $AC<0$ .

Given:

The equation, $A{x}^2+C{y}^2+Dx+Ey+F=0$ .

Concept used:

The equation of a hyperbola is given as $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ or $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ .

Note that in either case, the product of coefficients of ${\left(x-h\right)}^2$ and ${\left(y-k\right)}^2$ is $-\frac{1}{a^2{b}^2}<0$ , that is, negative.

Calculation:

The given equation is $A{x}^2+C{y}^2+Dx+Ey+F=0$ .

Simplifying,

  $\begin{array}{c}A{x}^2+C{y}^2+Dx+Ey+F=0\\ \left(A{x}^2+Dx\right)+\left(C{y}^2+Ey\right)+F=0\\ A\left(x^2+\frac{D}{A}x\right)+C\left(y^2+\frac{E}{C}y\right)+F=0\\ A\left(x^2+2\left(x\right)\left(\frac{D}{2A}\right)\right)+C\left(y^2+2\left(y\right)\left(\frac{E}{2C}\right)\right)+F=0\end{array}$

On further simplification,

  $\begin{array}{c}A\left(x^2+2\left(x\right)\left(\frac{D}{2A}\right)+{\left(\frac{D}{2A}\right)}^2-{\left(\frac{D}{2A}\right)}^2\right)+C\left(y^2+2\left(y\right)\left(\frac{E}{2C}\right)+{\left(\frac{E}{2C}\right)}^2-{\left(\frac{E}{2C}\right)}^2\right)+F=0\\ A\left({\left(x+\frac{D}{2A}\right)}^2-\frac{D^2}{4A^2}\right)+C\left({\left(y+\frac{E}{2C}\right)}^2-\frac{E^2}{4C^2}\right)+F=0\\ A{\left(x+\frac{D}{2A}\right)}^2-\frac{D^2}{4A}+C{\left(y+\frac{E}{2C}\right)}^2-\frac{E^2}{4C}+F=0\\ A{\left(x+\frac{D}{2A}\right)}^2+C{\left(y+\frac{E}{2C}\right)}^2=\frac{D^2}{4A}+\frac{E^2}{4C}-F\end{array}$

Continuing simplification,

  $\begin{array}{c}A{\left(x+\frac{D}{2A}\right)}^2+C{\left(y+\frac{E}{2C}\right)}^2=\frac{D^2}{4A}+\frac{E^2}{4C}-F\\ \frac{A{\left(x+\frac{D}{2A}\right)}^2+C{\left(y+\frac{E}{2C}\right)}^2}{\frac{D^2}{4A}+\frac{E^2}{4C}-F}=1\\ \frac{A{\left(x+\frac{D}{2A}\right)}^2}{\frac{D^2}{4A}+\frac{E^2}{4C}-F}+\frac{C{\left(y+\frac{E}{2C}\right)}^2}{\frac{D^2}{4A}+\frac{E^2}{4C}-F}=1\\ \frac{{\left(x+\frac{D}{2A}\right)}^2}{\left(\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{A}\right)}+\frac{{\left(y+\frac{E}{2C}\right)}^2}{\left(\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{C}\right)}=1\end{array}$

Now, the product of coefficients of ${\left(x+\frac{D}{2A}\right)}^2$ and ${\left(y+\frac{E}{2C}\right)}^2$ is $\frac{1}{\left(\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{A}\right)\left(\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{C}\right)}$ .

Now, as discussed, this is a hyperbola when $\frac{1}{\left(\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{A}\right)\left(\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{C}\right)}<0$ .

Simplifying,

  $\frac{1}{\frac{{\left(\frac{D^2}{4A}+\frac{E^2}{4C}-F\right)}^2}{AC}}<0$

On further simplification,

  $\frac{AC}{{\left(\frac{D^2}{4A}+\frac{E^2}{4C}-F\right)}^2}<0$

Since the denominator is always positive by the virtue of being a square quantity, the whole fraction is negative if and only if the numerator is negative.

Hence, $AC<0$ .

Thus, a non-degenerate graph of the given equation is a hyperbola if $AC<0$ .

Conclusion:

It has been shown that a non-degenerate graph of the given equation is a hyperbola if $AC<0$ .