The equation of the hyperbola with centre at (0, 0) and co-ordinate axes as its axes, distance between the directrices being `(4)/(sqrt(3))` and passing through the point (2, 1), is

Options

A. `3x^(2) + 2y^(2) = 10`
B. `3x^(2) - 2y^(2) = 2`
C. `2x^(2) - 3y^(2) =10`
D. `x^(2) -y^(2) =3`

Answer

To find the equation of the hyperbola with the given conditions, we can follow these steps:

### Step 1: Understand the properties of the hyperbola
The hyperbola is centered at (0, 0) and has the coordinate axes as its axes. Therefore, its standard form is given by:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Use the distance between the directrices
The distance between the directrices of a hyperbola is given by \( \frac{2a}{e} \), where \( e \) is the eccentricity. According to the problem, this distance is \( \frac{4}{\sqrt{3}} \).

Setting up the equation:

\[ \frac{2a}{e} = \frac{4}{\sqrt{3}} \]

From this, we can express \( \frac{a}{e} \):

\[ \frac{a}{e} = \frac{2}{\sqrt{3}} \]

### Step 3: Relate \( b^2 \) to \( a^2 \) and \( e \)
The relationship between \( a \), \( b \), and \( e \) for hyperbolas is given by:

\[ b^2 = a^2(e^2 - 1) \]

### Step 4: Substitute \( e \) in terms of \( a \)
From the earlier step, we have \( e = \frac{a \sqrt{3}}{2} \). Now substituting this into the equation for \( b^2 \):

\[ b^2 = a^2\left(\left(\frac{a \sqrt{3}}{2}\right)^2 - 1\right) \]

Calculating \( e^2 \):

\[ e^2 = \frac{3a^2}{4} \]

Thus,

\[ b^2 = a^2\left(\frac{3a^2}{4} - 1\right) = a^2\left(\frac{3a^2 - 4}{4}\right) = \frac{a^2(3a^2 - 4)}{4} \]

### Step 5: Use the point (2, 1) to find \( a \) and \( b \)
Since the hyperbola passes through the point (2, 1), we substitute \( x = 2 \) and \( y = 1 \) into the hyperbola equation:

\[ \frac{2^2}{a^2} - \frac{1^2}{b^2} = 1 \]

This simplifies to:

\[ \frac{4}{a^2} - \frac{1}{b^2} = 1 \]

Substituting \( b^2 \):

\[ \frac{4}{a^2} - \frac{4}{a^2(3a^2 - 4)} = 1 \]

### Step 6: Solve the equation
Multiplying through by \( a^2(3a^2 - 4) \):

\[ 4(3a^2 - 4) - 4 = a^2(3a^2 - 4) \]

This simplifies to:

\[ 12a^2 - 16 - 4 = 3a^4 - 4a^2 \]

Rearranging gives:

\[ 3a^4 - 16a^2 + 20 = 0 \]

### Step 7: Solve the quadratic equation
Let \( u = a^2 \):

\[ 3u^2 - 16u + 20 = 0 \]

Using the quadratic formula:

\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 3 \cdot 20}}{2 \cdot 3} \]

Calculating the discriminant:

\[ 256 - 240 = 16 \]

Thus,

\[ u = \frac{16 \pm 4}{6} \Rightarrow u = \frac{20}{6} = \frac{10}{3} \quad \text{or} \quad u = \frac{12}{6} = 2 \]

### Step 8: Find \( b^2 \) for both cases
1. For \( a^2 = \frac{10}{3} \):

\[ b^2 = \frac{10}{3} \left( \frac{3 \cdot \frac{10}{3}}{4} - 1 \right) = \frac{10}{3} \left( \frac{30}{12} - 1 \right) = \frac{10}{3} \cdot \frac{18}{12} = \frac{15}{6} = \frac{5}{2} \]

2. For \( a^2 = 2 \):

\[ b^2 = 2 \left( \frac{3 \cdot 2}{4} - 1 \right) = 2 \left( \frac{6}{4} - 1 \right) = 2 \cdot \frac{2}{4} = 1 \]

### Step 9: Write the equations of the hyperbola
1. For \( a^2 = \frac{10}{3} \) and \( b^2 = \frac{5}{2} \):

\[ \frac{x^2}{\frac{10}{3}} - \frac{y^2}{\frac{5}{2}} = 1 \Rightarrow 2x^2 - 3y^2 = 10 \]

2. For \( a^2 = 2 \) and \( b^2 = 1 \):

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

### Conclusion
The equations of the hyperbola are:

1. \( 2x^2 - 3y^2 = 10 \)
2. \( x^2 - 2y^2 = 2 \)

Among the options provided, the correct answer is:

**Option 2: \( 2x^2 - 3y^2 = 10 \)**