To find the point on the ellipse \( x^2 + 2y^2 = 6 \) that is nearest to the line \( x - y = 7 \), we can follow these steps:
### Step 1: Rewrite the line equation
First, we can rewrite the line equation in slope-intercept form:
\[
x - y = 7 \implies y = x - 7
\]
This shows that the slope of the line is 1.
### Step 2: Find the slope of the tangent to the ellipse
The general equation of the ellipse is given by:
\[
x^2 + 2y^2 = 6
\]
To find the slope of the tangent to the ellipse at any point \( (h, k) \), we can implicitly differentiate the ellipse equation:
\[
2x + 4y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{2y}
\]
This gives us the slope of the tangent line at any point \( (h, k) \) on the ellipse.
### Step 3: Set the slopes equal
Since we want the tangent to the ellipse to be parallel to the line \( y = x - 7 \), we set the slopes equal:
\[
-\frac{h}{2k} = 1 \implies h = -2k
\]
### Step 4: Substitute into the ellipse equation
Now we substitute \( h = -2k \) into the ellipse equation:
\[
(-2k)^2 + 2k^2 = 6 \implies 4k^2 + 2k^2 = 6 \implies 6k^2 = 6 \implies k^2 = 1 \implies k = 1 \text{ or } k = -1
\]
### Step 5: Find corresponding \( h \) values
Using \( k = 1 \):
\[
h = -2(1) = -2 \implies (-2, 1)
\]
Using \( k = -1 \):
\[
h = -2(-1) = 2 \implies (2, -1)
\]
### Step 6: Determine which point is nearest to the line
We need to find which of the points \( (-2, 1) \) or \( (2, -1) \) is closer to the line \( x - y = 7 \). We can calculate the distance from each point to the line using the formula for the distance from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \):
\[
\text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
\]
For the line \( x - y - 7 = 0 \) (where \( A = 1, B = -1, C = -7 \)):
1. For point \( (-2, 1) \):
\[
\text{Distance} = \frac{|1(-2) - 1(1) - 7|}{\sqrt{1^2 + (-1)^2}} = \frac{|-2 - 1 - 7|}{\sqrt{2}} = \frac{|-10|}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}
\]
2. For point \( (2, -1) \):
\[
\text{Distance} = \frac{|1(2) - 1(-1) - 7|}{\sqrt{1^2 + (-1)^2}} = \frac{|2 + 1 - 7|}{\sqrt{2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}
\]
### Conclusion
The point \( (2, -1) \) is closer to the line than \( (-2, 1) \). Therefore, the point on the ellipse \( x^2 + 2y^2 = 6 \) that is nearest to the line \( x - y = 7 \) is:
\[
\boxed{(2, -1)}
\]
