Question

Show that the graph of an equation of the form

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

where $A$and $C$are of the same sign,

(a) is an ellipse if $\frac{D^2}{4A}+\frac{E^2}{4C}-F$is the same sign as $A$.

(b) is a point if$\frac{D^2}{4A}+\frac{E^2}{4C}-F=0$.

(c) contains no points if $\frac{D^2}{4A}+\frac{E^2}{4C}-F$is of opposite sign to$A$.

Short answer

Part (a) The equation of ellipse is $\frac{{\left(x+\frac{D}{2A}\right)}^2}{\frac{D^2}{4A^2}+\frac{E^2}{4AC}-\frac{F}{A}}+\frac{{\left(y+\frac{E}{2C}\right)}^2}{\frac{D^2}{4AC}+\frac{E^2}{4C^2}-\frac{F}{C}}=1$.

Part (b) The coordinates of the point are $x=-\frac{D}{2A}\mathrm{and}y=-\frac{E}{2C}$.

Part (c) The equation contain no points if $\frac{D^2}{4A}+\frac{E^2}{4C}-F$ is of opposite sign to A.

Step-by-step solution

Step 1. Write the given information.

The given information is:

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

Part (a) Step 1. Complete the squares of x and y.

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

Now divide both sides with $-F+\frac{D^2}{4A}+\frac{E^2}{4C}$,

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

Therefore the equation of an ellipse is$\frac{{\left(x+\frac{D}{2A}\right)}^2}{\frac{D^2}{4A^2}+\frac{E^2}{4AC}-\frac{F}{A}}+\frac{{\left(y+\frac{E}{2C}\right)}^2}{\frac{D^2}{4AC}+\frac{E^2}{4C^2}-\frac{F}{C}}=1$ with center at$\left(\frac{-D}{2A},\frac{-E}{2C}\right)$if sign of$\frac{D^2}{4A}+\frac{E^2}{4C}-F$is the same sign as A.

Part (b) Step 1. Show that the equation is a point if D24C+E24C-F=0.

From part (a) we got the equation as: $A{\left(x+\frac{D}{2A}\right)}^2+C{\left(y+\frac{E}{2C}\right)}^2=-F+\frac{D^2}{4A}+\frac{E^2}{4C}$

Now it is given that $\frac{D^2}{4A}+\frac{E^2}{4C}-F=0$, therefore

$A{\left(x+\frac{D}{2A}\right)}^2+C{\left(y+\frac{E}{2C}\right)}^2=0$

This equation now becomes a point with coordinates$x=-\frac{D}{2A}\mathrm{and}y=-\frac{E}{2C}$.

Part (c) Step 1. Show that the equation contains no points if D24A+E24C-F is of opposite sign to A.

If the sign of $\frac{D^2}{4A}+\frac{E^2}{4C}-F$ is the opposite to A,the equation would contain no solution and it would not contain any point.

The reason being the result is an equation that has no answers because one side is always negative and the other always positive.