An eppipse of eccentricity `(2sqrt2)/(3)` is inscribed in a circle and a point within the circle is chosen at random. The probability that this point lies outside the ellipse is

Options

A. `1//9`
B. `4//9`
C. `1//3`
D. `2//3`

Answer

To solve the problem of finding the probability that a randomly chosen point within a circle lies outside an inscribed ellipse with eccentricity \( \frac{2\sqrt{2}}{3} \), we can follow these steps:

### Step 1: Understand the relationship between eccentricity, semi-major axis, and semi-minor axis
The eccentricity \( e \) of an ellipse is given by the formula:
\[ e = \frac{c}{a} \]
where \( c \) is the distance from the center to the focus, and \( a \) is the semi-major axis. Given \( e = \frac{2\sqrt{2}}{3} \), we can express \( c \) in terms of \( a \):
\[ c = \frac{2\sqrt{2}}{3} a \]

### Step 2: Use the relationship between \( a \), \( b \), and \( c \)
For an ellipse, we have the relationship:
\[ a^2 = b^2 + c^2 \]
where \( b \) is the semi-minor axis. Substituting \( c \) from the previous step:
\[ a^2 = b^2 + \left(\frac{2\sqrt{2}}{3} a\right)^2 \]
This simplifies to:
\[ a^2 = b^2 + \frac{8}{9} a^2 \]
Rearranging gives:
\[ b^2 = a^2 - \frac{8}{9} a^2 = \frac{1}{9} a^2 \]
Thus, we find:
\[ b = \frac{1}{3} a \]

### Step 3: Determine the semi-major and semi-minor axes
Let’s assume \( a = 3k \) (since the ellipse is inscribed in a circle, we can scale by \( k \)):
\[ b = \frac{1}{3} (3k) = k \]

### Step 4: Find the area of the ellipse
The area \( A_{\text{ellipse}} \) of the ellipse is given by:
\[ A_{\text{ellipse}} = \pi a b = \pi (3k)(k) = 3\pi k^2 \]

### Step 5: Find the area of the circle
The radius of the circle is equal to the semi-major axis \( a = 3k \). Thus, the area \( A_{\text{circle}} \) of the circle is:
\[ A_{\text{circle}} = \pi R^2 = \pi (3k)^2 = 9\pi k^2 \]

### Step 6: Calculate the area outside the ellipse but inside the circle
The area outside the ellipse but inside the circle is:
\[ A_{\text{outside}} = A_{\text{circle}} - A_{\text{ellipse}} = 9\pi k^2 - 3\pi k^2 = 6\pi k^2 \]

### Step 7: Calculate the probability
The probability \( P \) that a randomly chosen point inside the circle lies outside the ellipse is:
\[ P = \frac{A_{\text{outside}}}{A_{\text{circle}}} = \frac{6\pi k^2}{9\pi k^2} = \frac{6}{9} = \frac{2}{3} \]

Thus, the probability that a randomly chosen point lies outside the ellipse is \( \frac{2}{3} \).

### Final Answer
The probability that a randomly chosen point within the circle lies outside the ellipse is \( \frac{2}{3} \).

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