The point on the ellipse `x^(2)+2y^(2)=6` which is nearest to the line `x+y=7` is

Options

A. (a)`((sqrt6)/(sqrt5),(sqrt3)/(sqrt5))`
B. (b)`(2,1)`
C. (c)`(-2, 1)`
D. (d)`(-2, -1)`

Explanation

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 ellipse equation
The equation of the ellipse is given as:
\[ x^2 + 2y^2 = 6 \]
We can rewrite it in standard form by dividing both sides by 6:
\[ \frac{x^2}{6} + \frac{y^2}{3} = 1 \]

### Step 2: Find the distance from a point on the ellipse to the line
The distance \( d \) from a point \( (x_1, y_1) \) to the line \( ax + by + c = 0 \) is given by:
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
For the line \( x + y = 7 \), we can rewrite it as:
\[ x + y - 7 = 0 \]
Here, \( a = 1, b = 1, c = -7 \).

### Step 3: Set up the distance function
The distance from a point \( (x, y) \) on the ellipse to the line is:
\[ d = \frac{|x + y - 7|}{\sqrt{1^2 + 1^2}} = \frac{|x + y - 7|}{\sqrt{2}} \]
We want to minimize this distance.

### Step 4: Substitute \( y \) in terms of \( x \)
From the ellipse equation, we can express \( y \) in terms of \( x \):
\[ 2y^2 = 6 - x^2 \implies y^2 = \frac{6 - x^2}{2} \implies y = \sqrt{\frac{6 - x^2}{2}} \]
We will consider the positive root since we are looking for points in the first quadrant.

### Step 5: Substitute \( y \) into the distance function
Now we substitute \( y \) into the distance function:
\[ d = \frac{|x + \sqrt{\frac{6 - x^2}{2}} - 7|}{\sqrt{2}} \]
To minimize \( d \), we can minimize \( |x + \sqrt{\frac{6 - x^2}{2}} - 7| \).

### Step 6: Differentiate and find critical points
To find the minimum, we can differentiate the function:
Let \( f(x) = x + \sqrt{\frac{6 - x^2}{2}} - 7 \).
We will find \( f'(x) \) and set it to zero to find critical points.

Calculating \( f'(x) \):
\[ f'(x) = 1 + \frac{1}{2\sqrt{\frac{6 - x^2}{2}}} \cdot \left(-\frac{2x}{2}\right) = 1 - \frac{x}{\sqrt{3 - \frac{x^2}{2}}} \]
Setting \( f'(x) = 0 \):
\[ 1 - \frac{x}{\sqrt{3 - \frac{x^2}{2}}} = 0 \implies \frac{x}{\sqrt{3 - \frac{x^2}{2}}} = 1 \implies x = \sqrt{3 - \frac{x^2}{2}} \]

### Step 7: Solve for \( x \)
Squaring both sides:
\[ x^2 = 3 - \frac{x^2}{2} \implies \frac{3x^2}{2} = 3 \implies x^2 = 2 \implies x = \sqrt{2} \]

### Step 8: Find corresponding \( y \)
Now substituting \( x = \sqrt{2} \) back into the ellipse equation to find \( y \):
\[ y = \sqrt{\frac{6 - (\sqrt{2})^2}{2}} = \sqrt{\frac{6 - 2}{2}} = \sqrt{2} \]

### Step 9: Conclusion
Thus, the point on the ellipse that is nearest to the line \( x + y = 7 \) is:
\[ (\sqrt{2}, \sqrt{2}) \]