The is a point P on the hyperbola `(x^(2))/(16)-(y^(2))/(6)=1` such that its distance from the right directrix is the average of its distance from the two foci. Then the x-coordinate of P is (a)`-64/5` (b)`-32/9` (c)`-64/9` (d)none of these

Options

A. `-64//5`
B. `-32//9`
C. `-64//9`
D. none of these

Explanation

To solve the problem step by step, we will follow the given conditions and use the properties of hyperbolas.

### Step 1: Identify the hyperbola and its parameters
The equation of the hyperbola is given as:

\[ \frac{x^2}{16} - \frac{y^2}{6} = 1 \]

From this equation, we can identify:
- \( a^2 = 16 \) → \( a = 4 \)
- \( b^2 = 6 \) → \( b = \sqrt{6} \)

### Step 2: Find the foci and directrices
The foci of the hyperbola are located at:

\[ (\pm c, 0) \quad \text{where } c = \sqrt{a^2 + b^2} = \sqrt{16 + 6} = \sqrt{22} \]

Thus, the foci are at:

\[ (\sqrt{22}, 0) \quad \text{and} \quad (-\sqrt{22}, 0) \]

The equations of the directrices are given by:

\[ x = \pm \frac{a}{e} \quad \text{where } e = \frac{c}{a} = \frac{\sqrt{22}}{4} \]

Calculating \( e \):

\[ e = \frac{\sqrt{22}}{4} \]

Thus, the right directrix is:

\[ x = \frac{4}{\frac{\sqrt{22}}{4}} = \frac{16}{\sqrt{22}} \]

### Step 3: Set up the distance equations
Let the point \( P \) on the hyperbola be \( (x, y) \). The distance from the right directrix is:

\[ d_{directrix} = x - \frac{16}{\sqrt{22}} \]

The distances from the foci to the point \( P \) are:

\[ d_{focus1} = \sqrt{(x - \sqrt{22})^2 + y^2} \]
\[ d_{focus2} = \sqrt{(x + \sqrt{22})^2 + y^2} \]

### Step 4: Average distance from the foci
The average distance from the two foci is:

\[ d_{avg} = \frac{d_{focus1} + d_{focus2}}{2} \]

### Step 5: Set up the equation based on the problem statement
According to the problem, the distance from the right directrix is equal to the average of the distances from the two foci:

\[ x - \frac{16}{\sqrt{22}} = \frac{\sqrt{(x - \sqrt{22})^2 + y^2} + \sqrt{(x + \sqrt{22})^2 + y^2}}{2} \]

### Step 6: Substitute \( y^2 \) using the hyperbola equation
From the hyperbola equation, we have:

\[ y^2 = 6\left(\frac{x^2}{16} - 1\right) = \frac{6x^2}{16} - 6 = \frac{3x^2}{8} - 6 \]

### Step 7: Substitute \( y^2 \) into the distance equations
Now, substitute \( y^2 \) into the distance equations and simplify. This will yield a quadratic equation in \( x \).

### Step 8: Solve the quadratic equation
After simplification, solve the quadratic equation to find the possible values of \( x \).

### Final Step: Identify the correct x-coordinate
After solving, we will find the x-coordinate of point \( P \).