A point on the ellipse `x^(2)/16 + y^(2)/9 = 1` at a distance equal to the mean of lengths of the semi - major and semi-minor axis from the centre, is

Options

A. `((2sqrt(91))/(7), (3sqrt(105))/(14))`
B. `((2sqrt(91))/(7), (-3sqrt(91))/(14))`
C. `((-2sqrt(105))/(7), (-3sqrt(91))/(14))`
D. `((-2sqrt(105))/(7), (sqrt(91))/(14))`

Explanation

To solve the problem step by step, we will follow the given information about the ellipse and calculate the required point.

### Step 1: Identify the semi-major and semi-minor axes
The given ellipse is:

\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]

Comparing this with the standard form of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify:

- \(a^2 = 16 \Rightarrow a = 4\) (semi-major axis)
- \(b^2 = 9 \Rightarrow b = 3\) (semi-minor axis)

### Step 2: Calculate the mean of the semi-major and semi-minor axes
The mean of the lengths of the semi-major and semi-minor axes is given by:

\[ \text{Mean} = \frac{a + b}{2} = \frac{4 + 3}{2} = \frac{7}{2} \]

### Step 3: Set up the coordinates of the point on the ellipse
A point on the ellipse can be represented in parametric form as:

\[ P(\theta) = (a \cos \theta, b \sin \theta) = (4 \cos \theta, 3 \sin \theta) \]

### Step 4: Calculate the distance from the center to the point
The distance \(d\) from the center (0, 0) to the point \(P\) is given by:

\[ d = \sqrt{(4 \cos \theta)^2 + (3 \sin \theta)^2} = \sqrt{16 \cos^2 \theta + 9 \sin^2 \theta} \]

### Step 5: Set the distance equal to the mean
According to the problem, this distance is equal to the mean we calculated:

\[ \sqrt{16 \cos^2 \theta + 9 \sin^2 \theta} = \frac{7}{2} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:

\[ 16 \cos^2 \theta + 9 \sin^2 \theta = \left(\frac{7}{2}\right)^2 = \frac{49}{4} \]

### Step 7: Multiply through by 4 to eliminate the fraction
Multiplying through by 4 gives:

\[ 64 \cos^2 \theta + 36 \sin^2 \theta = 49 \]

### Step 8: Rearranging the equation
Rearranging gives:

\[ 64 \cos^2 \theta + 36 \sin^2 \theta - 49 = 0 \]

### Step 9: Substitute \(\sin^2 \theta = 1 - \cos^2 \theta\)
Substituting \(\sin^2 \theta = 1 - \cos^2 \theta\):

\[ 64 \cos^2 \theta + 36(1 - \cos^2 \theta) - 49 = 0 \]

This simplifies to:

\[ 64 \cos^2 \theta + 36 - 36 \cos^2 \theta - 49 = 0 \]

\[ (64 - 36) \cos^2 \theta - 13 = 0 \]

\[ 28 \cos^2 \theta = 13 \]

### Step 10: Solve for \(\cos^2 \theta\)
Thus,

\[ \cos^2 \theta = \frac{13}{28} \]

### Step 11: Find \(\sin^2 \theta\)
Using \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ \sin^2 \theta = 1 - \frac{13}{28} = \frac{15}{28} \]

### Step 12: Calculate the coordinates of the point
Now substituting back to find the coordinates:

\[ x = 4 \cos \theta = 4 \sqrt{\frac{13}{28}} = \frac{4 \sqrt{13}}{\sqrt{28}} = \frac{4 \sqrt{13}}{2\sqrt{7}} = \frac{2 \sqrt{13}}{\sqrt{7}} \]

\[ y = 3 \sin \theta = 3 \sqrt{\frac{15}{28}} = 3 \frac{\sqrt{15}}{\sqrt{28}} = \frac{3 \sqrt{15}}{2\sqrt{7}} \]

### Final Answer
Thus, the coordinates of the point on the ellipse at the specified distance from the center are:

\[ \left( \pm \frac{2 \sqrt{13}}{\sqrt{7}}, \pm \frac{3 \sqrt{15}}{2\sqrt{7}} \right) \]