Chapter 10.3, Problem 83AYU

a.

To determine

To show: An equation of the form $A{x}^2+C{y}^2+F=0$ is the equation of an ellipse with center at $\left(0,0\right)$ where $A\ne C$ .

Answer to Problem 83AYU

  $\frac{x^2}{C}+\frac{y^2}{A}=\frac{-F}{AC}$ is the equation of ellipse.

Explanation of Solution

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

Where $A\text{ and }C$ are of the same sign and $F$ is of opposite sign.

Yes, It is given that the $A{x}^2+C{y}^2+F=0,A\ne 0,C\ne 0,F\ne 0$

Where $A\text{ and }C$ are of the same sign and $F$ is of opposite sign.

Now

  $\begin{array}{l}A{x}^2+C{y}^2+F=0,A\ne 0,C\ne 0,F\ne 0\\ \frac{A{x}^2}{AC}+\frac{C{y}^2}{AC}=\frac{-F}{AC}\\ \frac{x^2}{C}+\frac{y^2}{A}=\frac{-F}{AC}\end{array}$

Hence, $\frac{x^2}{C}+\frac{y^2}{A}=\frac{-F}{AC}$ is the equation of ellipse.

b.

To determine

To show: An equation of the form $A{x}^2+C{y}^2+F=0$ is the equation of a circle with center at $\left(0,0\right)$ where $A=C$ .

Answer to Problem 83AYU

  $\frac{x^2}{A}+\frac{y^2}{A}=\frac{-F}{A^2}$ is the equation of circle.

Explanation of Solution

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

Where $A\text{ and }C$ are of the same sign and $F$ is of opposite sign.

Yes, It is given that the $A{x}^2+C{y}^2+F=0,A\ne 0,C\ne 0,F\ne 0$

Where $A\text{ and }C$ are of the same sign and $F$ is of opposite sign.

Now

  $\begin{array}{l}A{x}^2+C{y}^2+F=0,A\ne 0,C\ne 0,F\ne 0\\ \frac{A{x}^2}{AC}+\frac{C{y}^2}{AC}=\frac{-F}{AC}\\ \frac{x^2}{C}+\frac{y^2}{A}=\frac{-F}{AC}\\ \frac{x^2}{A}+\frac{y^2}{A}=\frac{-F}{A^2}\end{array}$

Hence, $\frac{x^2}{A}+\frac{y^2}{A}=\frac{-F}{A^2}$ is the equation of circle.