Find the equations of tangents to the ellipse `2x^2+y^2=8` which are
which makes an angle `pi/4` with x-axis.

Answer

To find the equations of tangents to the ellipse \(2x^2 + y^2 = 8\) that make an angle of \(\frac{\pi}{4}\) with the x-axis, we can follow these steps:

### Step 1: Write the equation of the ellipse in standard form
We start with the given equation of the ellipse:
\[ 2x^2 + y^2 = 8 \]
We divide the entire equation by 8 to convert it into standard form:
\[ \frac{2x^2}{8} + \frac{y^2}{8} = 1 \]
This simplifies to:
\[ \frac{x^2}{4} + \frac{y^2}{8} = 1 \]
Here, we identify \(a^2 = 4\) and \(b^2 = 8\).

### Step 2: Determine the slope of the tangent line
The angle given is \(\frac{\pi}{4}\). The slope \(m\) of the tangent line can be found using:
\[ m = \tan\left(\frac{\pi}{4}\right) = 1 \]

### Step 3: Write the equation of the tangent line
The general equation of a tangent line to the ellipse at slope \(m\) is given by:
\[ y = mx + c \]
Substituting \(m = 1\), we have:
\[ y = x + c \]

### Step 4: Find the value of \(c\)
To find \(c\), we use the formula:
\[ c = \pm \sqrt{a^2 m^2 + b^2} \]
Substituting \(a^2 = 4\), \(b^2 = 8\), and \(m = 1\):
\[ c = \pm \sqrt{4 \cdot 1^2 + 8} = \pm \sqrt{4 + 8} = \pm \sqrt{12} = \pm 2\sqrt{3} \]

### Step 5: Write the final equations of the tangents
Now substituting the values of \(c\) back into the equation of the tangent line, we get two equations:
1. \(y = x + 2\sqrt{3}\)
2. \(y = x - 2\sqrt{3}\)

Thus, the equations of the tangents to the ellipse \(2x^2 + y^2 = 8\) that make an angle of \(\frac{\pi}{4}\) with the x-axis are:
\[ y = x + 2\sqrt{3} \quad \text{and} \quad y = x - 2\sqrt{3} \]

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