Question

Show that an equation of the form

$A{x}^2+C{y}^2+F=0,A\ne 0,C\ne 0,F\ne 0$

where A and C are of the same sign and F is of opposite sign,

(a) is the equation of an ellipse with center at $\left(0,0\right)$if

$A\ne C$.

(b) is the equation of a circle with center $\left(0,0\right)$if $A=C$.

Short answer

Part (a) The equation of the ellipse with center at $\left(0,0\right)$is$\frac{x^2}{\left({\displaystyle \frac{-F}{A}}\right)}+\frac{y^2}{\left({\displaystyle \frac{-F}{C}}\right)}=1$.

Part (b) The equation of the circle with center $\left(0,0\right)$is$x^2+y^2=-\frac{F}{A}$.

Step-by-step solution

Step 1. Write the given information.

The given information is:

Show that an equation of the form

$A{x}^2+C{y}^2+F=0,A\ne 0,C\ne 0,F\ne 0$

Part (a) Step 1. Subtract F from both sides then divide both sides by -F.

Subtract F from both the sides,

$$\begin{gathered} A{x}^2+C{y}^2+F=0 \\ A{x}^2+C{y}^2+F-F=0-F \\ A{x}^2+C{y}^2=-F \end{gathered}$$

Now divide both sides by $-F$,

$$\begin{gathered} \frac{A{x}^2}{-F}+\frac{C{y}^2}{-F}=\frac{-F}{-F} \\ \frac{x^2}{\left({\displaystyle \frac{-F}{A}}\right)}+\frac{y^2}{\left({\displaystyle \frac{-F}{C}}\right)}=1 \end{gathered}$$

Part (a) Step 2. Write the equation of the ellipse with center at 0,0.

As we can see that that $A\mathrm{and}F\mathrm{and}C\mathrm{and}F$have opposite signs, the denominator of the equation $-\frac{F}{A}\mathrm{and}-\frac{F}{C}$will be positive.

$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$is the general form of equation of an ellipse.

Therefore, the equation of an ellipse with center at$\left(0,0\right)$is$\frac{x^2}{\left({\displaystyle \frac{-F}{A}}\right)}+\frac{y^2}{\left({\displaystyle \frac{-F}{C}}\right)}=1$.

Part (b) Step 1. Subtract F from both sides of the given equation then divide both sides by A.

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

We can rewrite the given equation as $A{x}^2+A{y}^2+F=0$because it is given that $A=C$.

Subtract $F$from both sides,

$$\begin{gathered} A{x}^2+A{y}^2+F=0 \\ A{x}^2+A{y}^2+F-F=0-F \\ A{x}^2+A{y}^2=-F \end{gathered}$$

Now divide both sides by $A$,

$$\begin{gathered} \frac{A{x}^2}{A}+\frac{A{y}^2}{A}=\frac{-F}{A} \\ {x}^2+y^2=-\frac{F}{A} \end{gathered}$$

Part (b) Step 2. Find the equation of the circle with center at 0,0.

As we can see that that $A\mathrm{and}F$have opposite signs, the fraction $-\frac{F}{A}$ will be positive.

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$is the equation of the circle.

Then the equation$x^2+y^2=-\frac{F}{A}$is the form of circle with radius$\sqrt{-\frac{F}{A}}$and center at$\left(0,0\right)$.