To find the distance from the center of the ellipse given by the equation \( x^2 + 2y^2 - 2 = 0 \) to the tangents of the ellipse that are equally inclined to both axes, we can follow these steps:
### Step 1: Rewrite the equation of the ellipse
The given equation of the ellipse is:
\[
x^2 + 2y^2 - 2 = 0
\]
We can rewrite this as:
\[
x^2 + 2y^2 = 2
\]
Dividing through by 2 gives:
\[
\frac{x^2}{2} + \frac{y^2}{1} = 1
\]
This is the standard form of the ellipse.
### Step 2: Identify the center of the ellipse
From the standard form of the ellipse, we can see that the center of the ellipse is at the point:
\[
(0, 0)
\]
### Step 3: Determine the slopes of the tangents
The tangents that are equally inclined to both axes will have slopes of \( m = 1 \) and \( m = -1 \) (which correspond to angles of 45° and 135° with respect to the x-axis).
### Step 4: Write the equations of the tangents
The general equation of the tangent to the ellipse at a slope \( m \) is given by:
\[
y = mx \pm \sqrt{2m^2 + 1}
\]
For \( m = 1 \):
\[
y = x \pm \sqrt{2(1)^2 + 1} = x \pm \sqrt{3}
\]
Thus, the equations of the tangents for \( m = 1 \) are:
1. \( y = x + \sqrt{3} \)
2. \( y = x - \sqrt{3} \)
For \( m = -1 \):
\[
y = -x \pm \sqrt{2(-1)^2 + 1} = -x \pm \sqrt{3}
\]
Thus, the equations of the tangents for \( m = -1 \) are:
3. \( y = -x + \sqrt{3} \)
4. \( y = -x - \sqrt{3} \)
### Step 5: Calculate the distance from the center to the tangents
To find the distance from the center (0, 0) to the tangent lines, we can use the distance formula from a point to a line given by \( Ax + By + C = 0 \):
\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]
For the tangent lines, we can rewrite them in the form \( Ax + By + C = 0 \).
1. For \( y = x + \sqrt{3} \):
- Rearranging gives: \( x - y + \sqrt{3} = 0 \)
- Here, \( A = 1, B = -1, C = \sqrt{3} \)
The distance from (0, 0) is:
\[
d = \frac{|1(0) - 1(0) + \sqrt{3}|}{\sqrt{1^2 + (-1)^2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2}
\]
2. For \( y = x - \sqrt{3} \):
- Rearranging gives: \( x - y - \sqrt{3} = 0 \)
- Here, \( A = 1, B = -1, C = -\sqrt{3} \)
The distance from (0, 0) is:
\[
d = \frac{|1(0) - 1(0) - \sqrt{3}|}{\sqrt{1^2 + (-1)^2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2}
\]
3. For \( y = -x + \sqrt{3} \):
- Rearranging gives: \( x + y - \sqrt{3} = 0 \)
- Here, \( A = 1, B = 1, C = -\sqrt{3} \)
The distance from (0, 0) is:
\[
d = \frac{|1(0) + 1(0) - \sqrt{3}|}{\sqrt{1^2 + 1^2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2}
\]
4. For \( y = -x - \sqrt{3} \):
- Rearranging gives: \( x + y + \sqrt{3} = 0 \)
- Here, \( A = 1, B = 1, C = \sqrt{3} \)
The distance from (0, 0) is:
\[
d = \frac{|1(0) + 1(0) + \sqrt{3}|}{\sqrt{1^2 + 1^2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2}
\]
### Conclusion
The distance from the center of the ellipse to each of the tangents that are equally inclined to both axes is:
\[
\frac{\sqrt{6}}{2}
\]
