If the normal from the point P(h,1) on the ellipse `x^2/6+y^2/3=1` is perpendicular to the line `x+y=8` , then the value of h is

Options

A. 1
B. 2
C. 8
D. 9

Explanation

To solve the problem step-by-step, we will follow the given information and use the properties of the ellipse and the concept of slopes.

### Step 1: Understand the given ellipse
The equation of the ellipse is given as:
\[ \frac{x^2}{6} + \frac{y^2}{3} = 1 \]
From this, we can identify:
- \(a^2 = 6\)
- \(b^2 = 3\)

### Step 2: Write the equation of the normal to the ellipse
For any point \((x_1, y_1)\) on the ellipse, the equation of the normal can be expressed as:
\[ \frac{a^2 x_1}{a^2} x + \frac{b^2 y_1}{b^2} y = a^2 - b^2 \]
Substituting \(x_1 = h\) and \(y_1 = 1\):
\[ \frac{6h}{6} x - \frac{3 \cdot 1}{3} y = 6 - 3 \]
This simplifies to:
\[ x - y = 1 \]
or rearranging gives:
\[ y = x - 1 \]

### Step 3: Find the slope of the normal
From the equation \(y = x - 1\), we can see that the slope of the normal line is:
\[ m_{\text{normal}} = 1 \]

### Step 4: Analyze the given line
The line given is:
\[ x + y = 8 \]
Rearranging gives:
\[ y = -x + 8 \]
The slope of this line is:
\[ m_{\text{line}} = -1 \]

### Step 5: Use the property of perpendicular lines
For two lines to be perpendicular, the product of their slopes must equal -1. Thus, we have:
\[ m_{\text{normal}} \times m_{\text{line}} = -1 \]
Substituting the slopes:
\[ 1 \times (-1) = -1 \]
This confirms that the normal is indeed perpendicular to the line.

### Step 6: Set up the equation for the normal's slope
From the normal equation derived earlier, we can express it in slope-intercept form:
\[ y = \frac{2}{h} x - 1 \]
Thus, the slope of the normal can also be expressed as:
\[ m_{\text{normal}} = \frac{2}{h} \]

### Step 7: Set the slopes equal to find \(h\)
Since we have established that the slope of the normal is 1, we can set up the equation:
\[ \frac{2}{h} = 1 \]
Solving for \(h\):
\[ 2 = h \implies h = 2 \]

### Conclusion
The value of \(h\) is:
\[ \boxed{2} \]