What is the ratio of the greatest and least focal distances ofa point on the ellipse `4x^(2)+9y^(2)=36?`

Explanation

To find the ratio of the greatest and least focal distances of a point on the ellipse given by the equation \(4x^2 + 9y^2 = 36\), we will follow these steps:

### Step 1: Rewrite the equation of the ellipse in standard form.
The given equation is:
\[ 4x^2 + 9y^2 = 36 \]
Dividing through by 36, we get:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
This is now in the standard form of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).

### Step 2: Identify the values of \(a\) and \(b\).
From the standard form, we can identify:
\[ a^2 = 9 \quad \Rightarrow \quad a = 3 \]
\[ b^2 = 4 \quad \Rightarrow \quad b = 2 \]

### Step 3: Calculate the eccentricity \(e\) of the ellipse.
The eccentricity \(e\) is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values of \(a\) and \(b\):
\[ e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Step 4: Find the greatest and least focal distances.
The greatest focal distance \(d_{max}\) and least focal distance \(d_{min}\) are given by:
\[ d_{max} = a + ae = a(1 + e) \]
\[ d_{min} = a - ae = a(1 - e) \]
Substituting the values of \(a\) and \(e\):
\[ d_{max} = 3(1 + \frac{\sqrt{5}}{3}) = 3 + \sqrt{5} \]
\[ d_{min} = 3(1 - \frac{\sqrt{5}}{3}) = 3 - \sqrt{5} \]

### Step 5: Calculate the ratio of the greatest to the least focal distances.
The ratio \(R\) of the greatest to the least focal distances is:
\[ R = \frac{d_{max}}{d_{min}} = \frac{3 + \sqrt{5}}{3 - \sqrt{5}} \]

### Step 6: Simplify the ratio.
To simplify this ratio, we can multiply the numerator and denominator by the conjugate of the denominator:
\[ R = \frac{(3 + \sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} \]
Calculating the denominator:
\[ (3 - \sqrt{5})(3 + \sqrt{5}) = 9 - 5 = 4 \]
Calculating the numerator:
\[ (3 + \sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5} \]
Thus, we have:
\[ R = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2} \]

### Final Answer
The ratio of the greatest and least focal distances of a point on the ellipse is:
\[ \frac{7 + 3\sqrt{5}}{2} \]