The equation of the tangents to the hyperbola `3x^(2) -4y^(2) =12 ` which are parallel to the line ` 2x+ y+7=0 ` are

Options

A. A :`2x+y pm sqrt13=0`
B. B: `3x-y pm sqrt16=0`
C. C: `3x-y pm 2=0`
D. D:`3x-y pm sqrt2=0`

Answer

To find the equations of the tangents to the hyperbola \(3x^2 - 4y^2 = 12\) that are parallel to the line \(2x + y + 7 = 0\), we can follow these steps:

### Step 1: Rewrite the hyperbola in standard form
The given equation of the hyperbola is:
\[ 3x^2 - 4y^2 = 12 \]
To rewrite it in standard form, divide the entire equation by 12:
\[ \frac{x^2}{4} - \frac{y^2}{3} = 1 \]
This shows that \(a^2 = 4\) and \(b^2 = 3\).

### Step 2: Determine the slope of the given line
The equation of the line is:
\[ 2x + y + 7 = 0 \]
We can rewrite it in slope-intercept form \(y = mx + c\):
\[ y = -2x - 7 \]
From this, we see that the slope \(m\) of the line is \(-2\).

### Step 3: Use the slope to find the tangent equation
The equation of the tangent to the hyperbola in slope-intercept form is given by:
\[ y = mx \pm \sqrt{a^2 m^2 - b^2} \]
Substituting \(m = -2\), \(a^2 = 4\), and \(b^2 = 3\):
\[ y = -2x \pm \sqrt{4(-2)^2 - 3} \]
Calculating the expression under the square root:
\[ 4(-2)^2 = 4 \times 4 = 16 \]
So,
\[ y = -2x \pm \sqrt{16 - 3} = -2x \pm \sqrt{13} \]

### Step 4: Write the equations of the tangents
Thus, the equations of the tangents can be expressed as:
\[ y + 2x \pm \sqrt{13} = 0 \]
This can be rewritten as:
\[ 2x + y + \sqrt{13} = 0 \quad \text{and} \quad 2x + y - \sqrt{13} = 0 \]

### Final Answer
The equations of the tangents to the hyperbola \(3x^2 - 4y^2 = 12\) which are parallel to the line \(2x + y + 7 = 0\) are:
\[ 2x + y + \sqrt{13} = 0 \quad \text{and} \quad 2x + y - \sqrt{13} = 0 \]

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