Chapter 10.5, Problem 8AYU

To determine

The choice in the blank for the following statement

The equation $a{x}^2+6y^2+12y=0$ defines an ellipse if $\underset{\_}{}$

1. $a<0$

2. $a=0$

3. $a>0$

4. $a$ is any real number.

Answer to Problem 8AYU

Solution:

The equation $a{x}^2+6y^2+12y=0$ defines an ellipse, if $a>0$

Explanation of Solution

Given information:

The equation is $a{x}^2+6y^2+12y=0$

Explanation:

The equation $A{x}^2+C{y}^2+Dx+Ey+F=0$, where $A$ and $C$ are not both equal to zero defines an ellipse, if $AC>0$

Compare the equation $a{x}^2+6y^2+12y=0$ and $a{x}^2+6y^2+12y=0$, get

$A=a$ and $C=6$

Implies that $AC=6a$

Since the equation defines a ellipse, when $AC>0$, that is when $6a>0$

Dividing both sides of $6a>0$ by $6$

$\Rightarrow a>0$

Therefore, the equation $a{x}^2+6y^2+12y=0$ defines an ellipse if $a>0$