Equations of tangents to the hyperbola `4x^(2)-3y^(2)=24` which makes an angle `30^(@)` with y-axis are

Options

A. `sqrt3x+y= pm sqrt10`
B. `sqrt3x-y= pm 10`
C. `sqrt3x-y= pm 5`
D. `sqrt3x-y= pm sqrt5`

Answer

To find the equations of the tangents to the hyperbola \(4x^2 - 3y^2 = 24\) that make an angle of \(30^\circ\) with the y-axis, we can follow these steps:

### Step 1: Rewrite the Hyperbola in Standard Form
We start with the given equation of the hyperbola:
\[ 4x^2 - 3y^2 = 24 \]
Dividing both sides by 24, we get:
\[ \frac{x^2}{6} - \frac{y^2}{8} = 1 \]
This is now in the standard form of a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a^2 = 6\) and \(b^2 = 8\).

### Step 2: Identify \(a\) and \(b\)
From the standard form, we can identify:
\[ a = \sqrt{6}, \quad b = 2\sqrt{2} \]

### Step 3: Find the Slope of the Tangent
Since the tangent makes an angle of \(30^\circ\) with the y-axis, we need to find the slope \(m\) of the tangent line. The angle with the x-axis will be:
\[ 90^\circ - 30^\circ = 60^\circ \]
Thus, the slope \(m\) is:
\[ m = \tan(60^\circ) = \sqrt{3} \]

### Step 4: Use the Tangent Equation
The equation of the tangent to the hyperbola can be expressed as:
\[ y = mx + \sqrt{a^2 m^2 - b^2} \]
Substituting \(m = \sqrt{3}\), \(a^2 = 6\), and \(b^2 = 8\):
\[ y = \sqrt{3}x \pm \sqrt{6 \cdot 3 - 8} \]

### Step 5: Calculate the Constant Term
Calculating the term under the square root:
\[ 6 \cdot 3 - 8 = 18 - 8 = 10 \]
Thus, we have:
\[ y = \sqrt{3}x \pm \sqrt{10} \]

### Step 6: Rearranging the Equation
Rearranging the equation gives us:
\[ \sqrt{3}x - y = \mp \sqrt{10} \]
This leads to two equations:
1. \(\sqrt{3}x - y = \sqrt{10}\)
2. \(\sqrt{3}x - y = -\sqrt{10}\)

### Conclusion
The equations of the tangents to the hyperbola that make an angle of \(30^\circ\) with the y-axis are:
\[ \sqrt{3}x - y = \pm \sqrt{10} \]