The equations of the tangents to the ellipse `3x^(2)+4y^(2)=12` which are parallel to the line `2x-y+5=0` is

Options

A. `6x-2ypmsqrt(155//3)=0`
B. `2x-ypmsqrt(19)=0`
C. `16x+22ypmsqrt(155//3)=0`
D. `2x+2ypmsqrt(39)=1`

Answer

To find the equations of the tangents to the ellipse \(3x^2 + 4y^2 = 12\) that are parallel to the line \(2x - y + 5 = 0\), we can follow these steps:

### Step 1: Rewrite the equation of the ellipse in standard form
We start with the equation of the ellipse:
\[ 3x^2 + 4y^2 = 12 \]
Dividing both sides by 12 gives:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]
This shows that \(a^2 = 4\) and \(b^2 = 3\), where \(a = 2\) and \(b = \sqrt{3}\).

### Step 2: Identify the slope of the given line
The equation of the line is:
\[ 2x - y + 5 = 0 \]
Rearranging this into slope-intercept form \(y = mx + c\):
\[ y = 2x + 5 \]
From this, we can see that the slope \(m\) of the line is \(2\).

### Step 3: Use the formula for the tangent to the ellipse
For an ellipse in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the equation of the tangent line with slope \(m\) is given by:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
Substituting \(m = 2\), \(a^2 = 4\), and \(b^2 = 3\) into the formula:
\[ y = 2x \pm \sqrt{4 \cdot 2^2 + 3} \]
Calculating the expression under the square root:
\[ 4 \cdot 4 + 3 = 16 + 3 = 19 \]
Thus, we have:
\[ y = 2x \pm \sqrt{19} \]

### Step 4: Write the equations of the tangents
The equations of the tangents can be written as:
\[ y = 2x + \sqrt{19} \quad \text{and} \quad y = 2x - \sqrt{19} \]
Rearranging these into standard form gives:
\[ 2x - y + \sqrt{19} = 0 \quad \text{and} \quad 2x - y - \sqrt{19} = 0 \]

### Final Answer
The equations of the tangents to the ellipse that are parallel to the line \(2x - y + 5 = 0\) are:
\[ 2x - y + \sqrt{19} = 0 \quad \text{and} \quad 2x - y - \sqrt{19} = 0 \]