The distance of the centre of the ellipse `x^2 +2y^2 -2 = 0` to those tangents of the ellipse which are equally inclined to both the axes is

Options

A. `(3)/(sqrt2)`
B. `(sqrt3)/(2)`
C. `(sqrt2)/(3)`
D. `sqrt(3/2)`

Explanation

To solve the problem, we need to find the distance from the center of the ellipse \( x^2 + 2y^2 - 2 = 0 \) to the tangents of the ellipse that are equally inclined to both axes.

### Step 1: Rewrite the equation of the ellipse
We start with the given equation of the ellipse:
\[ x^2 + 2y^2 - 2 = 0 \]
Rearranging gives:
\[ x^2 + 2y^2 = 2 \]
Now, divide the entire equation by 2:
\[ \frac{x^2}{2} + \frac{y^2}{1} = 1 \]
This is the standard form of the ellipse.

### Step 2: Identify the center of the ellipse
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can see that the center of the ellipse is at the origin \((0, 0)\).

### Step 3: Determine the slope of the tangents
The problem states that we need to find tangents that are equally inclined to both axes. The slopes of such tangents are \(m = 1\) and \(m = -1\).

### Step 4: Write the equation of the tangent
The equation of the tangent to the ellipse at a slope \(m\) is given by:
\[ y = mx \pm \sqrt{2m^2 + 1} \]
Substituting \(m = 1\):
\[ y = x \pm \sqrt{2(1)^2 + 1} = x \pm \sqrt{3} \]
Substituting \(m = -1\):
\[ y = -x \pm \sqrt{2(-1)^2 + 1} = -x \pm \sqrt{3} \]
Thus, we have four tangent lines:
1. \(y = x + \sqrt{3}\)
2. \(y = x - \sqrt{3}\)
3. \(y = -x + \sqrt{3}\)
4. \(y = -x - \sqrt{3}\)

### Step 5: Calculate the distance from the center to the tangents
The distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the tangent \(y = x + \sqrt{3}\), we can rewrite it in the form \(Ax + By + C = 0\):
\[ x - y + \sqrt{3} = 0 \quad \Rightarrow \quad A = 1, B = -1, C = \sqrt{3} \]
Substituting \((x_1, y_1) = (0, 0)\):
\[ d = \frac{|1(0) - 1(0) + \sqrt{3}|}{\sqrt{1^2 + (-1)^2}} = \frac{|\sqrt{3}|}{\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} \]

### Step 6: Calculate for the other tangents
Repeating this for the other tangents, we will find that the distance remains the same due to symmetry.

### Final Answer
The distance from the center of the ellipse to the tangents that are equally inclined to both axes is:
\[ \frac{\sqrt{6}}{2} \]