Understanding the Equation of an Ellipse

The given equation is \(\frac{{{x^2}}}{{2 - r}} + \frac{{{y^2}}}{{r - 6}} + 1 = 0\). We want to find the conditions on the parameter \(r\) for this equation to represent an equation of an ellipse. An ellipse is a type of conic section.

Rewriting the Ellipse Equation

The standard form of an ellipse equation centered at the origin is \(\frac{{{x^2}}}{{a^2}} + \frac{{{y^2}}}{{b^2}} = 1\), where \(a^2\) and \(b^2\) are positive constants. Let's rewrite the given equation to match this standard form:

Start with the given equation:

\(\frac{{{x^2}}}{{2 - r}} + \frac{{{y^2}}}{{r - 6}} + 1 = 0\)

Subtract 1 from both sides:

\(\frac{{{x^2}}}{{2 - r}} + \frac{{{y^2}}}{{r - 6}} = -1\)

To get +1 on the right side, we can multiply the entire equation by -1, or equivalently, move the negative sign into the denominators:

\(\frac{{{x^2}}}{{-(2 - r)}} + \frac{{{y^2}}}{{-(r - 6)}} = 1\)

Simplify the denominators:

\(\frac{{{x^2}}}{{r - 2}} + \frac{{{y^2}}}{{6 - r}} = 1\)

Now the equation is in the standard form \(\frac{{{x^2}}}{{A}} + \frac{{{y^2}}}{{B}} = 1\), where \(A = r - 2\) and \(B = 6 - r\).

Conditions for an Ellipse

For the equation \(\frac{{{x^2}}}{{A}} + \frac{{{y^2}}}{{B}} = 1\) to represent an equation of an ellipse, the denominators \(A\) and \(B\) must be positive and distinct. If they were equal, it would be a circle (a special case of an ellipse). If either were negative or zero, it would represent a hyperbola, parabola, or degenerate conic section.

So, we need two conditions for ellipse:

• The first denominator must be positive: \(r - 2 > 0\)
• The second denominator must be positive: \(6 - r > 0\)

Determining the Range of Parameter r

Let's solve each inequality for the parameter \(r\):

1. From the first condition, \(r - 2 > 0\):
   Add 2 to both sides: \(r > 2\)
2. From the second condition, \(6 - r > 0\):
   Add \(r\) to both sides: \(6 > r\)
   This can also be written as: \(r < 6\)

For the equation to represent an equation of an ellipse, both conditions must hold true simultaneously. We need \(r > 2\) AND \(r < 6\).

Combining these two inequalities gives us the range for the parameter \(r\):

\(2 < r < 6\)

If \(r\) is within this range, both denominators \(r-2\) and \(6-r\) will be positive, satisfying the conditions for ellipse and ensuring the equation represents an ellipse.