Find the equations of the tangents to the ellipse `2x^(2) + y^(2) = 8` which are
(i) parallel to x - 2y - 4 = 0
(ii) perpendicular to x + y + 2 = 0

Answer

To find the equations of the tangents to the ellipse \(2x^2 + y^2 = 8\) that are (i) parallel to \(x - 2y - 4 = 0\) and (ii) perpendicular to \(x + y + 2 = 0\), we will follow these steps:

### Step 1: Rewrite the equation of the ellipse in standard form
The given equation of the ellipse is:
\[ 2x^2 + y^2 = 8 \]
Dividing both sides by 8, we get:
\[ \frac{x^2}{4} + \frac{y^2}{8} = 1 \]
This shows that \(a^2 = 4\) and \(b^2 = 8\).

### Step 2: Find the slope of the line parallel to the first tangent
The equation of the line is:
\[ x - 2y - 4 = 0 \]
Rearranging it to slope-intercept form \(y = mx + c\):
\[ 2y = x - 4 \implies y = \frac{1}{2}x - 2 \]
Thus, the slope \(m\) of the line is \(\frac{1}{2}\).

### Step 3: Use the formula for the tangent to the ellipse
The equation of the tangent to the ellipse at a point with slope \(m\) is given by:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
Substituting \(m = \frac{1}{2}\), \(a^2 = 4\), and \(b^2 = 8\):
\[ y = \frac{1}{2}x \pm \sqrt{4 \cdot \left(\frac{1}{2}\right)^2 + 8} \]
Calculating the expression under the square root:
\[ 4 \cdot \left(\frac{1}{2}\right)^2 = 4 \cdot \frac{1}{4} = 1 \]
Thus:
\[ y = \frac{1}{2}x \pm \sqrt{1 + 8} = \frac{1}{2}x \pm 3 \]
This gives us two equations:
\[ y = \frac{1}{2}x + 3 \quad \text{and} \quad y = \frac{1}{2}x - 3 \]

### Step 4: Convert to standard form
Multiplying both equations by 2 to eliminate the fraction:
1. For \(y = \frac{1}{2}x + 3\):
\[ 2y = x + 6 \implies x - 2y + 6 = 0 \]
2. For \(y = \frac{1}{2}x - 3\):
\[ 2y = x - 6 \implies x - 2y - 6 = 0 \]

### Step 5: Find the slope of the line perpendicular to the second tangent
The equation of the line is:
\[ x + y + 2 = 0 \]
Rearranging it to slope-intercept form:
\[ y = -x - 2 \]
Thus, the slope of this line is \(-1\). The slope of the tangent line that is perpendicular to this line will be the negative reciprocal:
\[ m = 1 \]

### Step 6: Use the formula for the tangent to the ellipse
Substituting \(m = 1\):
\[ y = x \pm \sqrt{4 \cdot 1^2 + 8} \]
Calculating the expression under the square root:
\[ 4 \cdot 1^2 = 4 \]
Thus:
\[ y = x \pm \sqrt{4 + 8} = x \pm \sqrt{12} = x \pm 2\sqrt{3} \]
This gives us two equations:
\[ y = x + 2\sqrt{3} \quad \text{and} \quad y = x - 2\sqrt{3} \]

### Step 7: Convert to standard form
Multiplying both equations by 1:
1. For \(y = x + 2\sqrt{3}\):
\[ y - x - 2\sqrt{3} = 0 \]
2. For \(y = x - 2\sqrt{3}\):
\[ y - x + 2\sqrt{3} = 0 \]

### Final Result
The equations of the tangents to the ellipse are:
1. \(x - 2y + 6 = 0\) and \(x - 2y - 6 = 0\) (for the first part)
2. \(y - x - 2\sqrt{3} = 0\) and \(y - x + 2\sqrt{3} = 0\) (for the second part)