If a directrix of a hyperbola centred at the origin and passing through the point `(4,-2 sqrt(3)) " is " 5x=4sqrt(5)` and its eccentricity is e, then

Explanation

To solve the problem step by step, we will follow the mathematical reasoning based on the information provided.

### Step 1: Identify the equation of the hyperbola
The standard form of a hyperbola centered at the origin is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Use the point on the hyperbola
The hyperbola passes through the point \( (4, -2\sqrt{3}) \). Substituting these coordinates into the hyperbola equation gives:
\[ \frac{4^2}{a^2} - \frac{(-2\sqrt{3})^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{16}{a^2} - \frac{12}{b^2} = 1 \]

### Step 3: Relate \( a \) and \( b \) using the eccentricity
The eccentricity \( e \) of a hyperbola is related to \( a \) and \( b \) by the equation:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
From this, we can express \( b^2 \) in terms of \( a^2 \) and \( e \):
\[ b^2 = a^2(e^2 - 1) \]

### Step 4: Substitute \( b^2 \) into the hyperbola equation
Substituting \( b^2 \) into the equation from Step 2 gives:
\[ \frac{16}{a^2} - \frac{12}{a^2(e^2 - 1)} = 1 \]

### Step 5: Simplify the equation
Multiply through by \( a^2(e^2 - 1) \) to eliminate the denominators:
\[ 16(e^2 - 1) - 12 = a^2(e^2 - 1) \]
This simplifies to:
\[ 16e^2 - 16 - 12 = a^2(e^2 - 1) \]
\[ 16e^2 - 28 = a^2(e^2 - 1) \]

### Step 6: Use the directrix equation
The equation of the directrix is given as \( 5x = 4\sqrt{5} \). The distance from the center to the directrix is:
\[ \frac{a}{e} = \frac{4\sqrt{5}}{5} \]
From this, we can express \( a \) in terms of \( e \):
\[ a = \frac{4\sqrt{5}}{5} e \]

### Step 7: Substitute \( a \) back into the equation
Substituting \( a \) into the equation from Step 5:
\[ 16e^2 - 28 = \left(\frac{4\sqrt{5}}{5} e\right)^2 (e^2 - 1) \]
Calculating \( a^2 \):
\[ \left(\frac{4\sqrt{5}}{5} e\right)^2 = \frac{16 \cdot 5}{25} e^2 = \frac{16}{5} e^2 \]
Substituting this into the equation gives:
\[ 16e^2 - 28 = \frac{16}{5} e^2 (e^2 - 1) \]

### Step 8: Clear the fractions
Multiply through by 5 to eliminate the fraction:
\[ 80e^2 - 140 = 16e^2(e^2 - 1) \]
Expanding the right side:
\[ 80e^2 - 140 = 16e^4 - 16e^2 \]

### Step 9: Rearrange the equation
Rearranging gives:
\[ 16e^4 - 96e^2 + 140 = 0 \]

### Step 10: Solve the quadratic equation
Let \( x = e^2 \). The equation becomes:
\[ 16x^2 - 96x + 140 = 0 \]
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{96 \pm \sqrt{(-96)^2 - 4 \cdot 16 \cdot 140}}{2 \cdot 16} \]

### Step 11: Calculate the discriminant and find \( e \)
Calculate the discriminant and solve for \( e \):
\[ x = \frac{96 \pm \sqrt{9216 - 8960}}{32} = \frac{96 \pm \sqrt{256}}{32} = \frac{96 \pm 16}{32} \]
This gives two potential values for \( x \):
\[ x_1 = \frac{112}{32} = 3.5, \quad x_2 = \frac{80}{32} = 2.5 \]
Thus, \( e^2 = 3.5 \) or \( e^2 = 2.5 \).

### Final Answer
The eccentricity \( e \) can be found as:
\[ e = \sqrt{3.5} \text{ or } e = \sqrt{2.5} \]