If a point `P` is taken on `xy=2` and then a normal is drawn from `P` on the ellipse `(x^(2))/6+(y^(2))/3=1` which is perpendicular to `x+y=8`, then `P` is

Options

A. `(1,2)`
B. `(-1,-2)`
C. `(2,1)`
D. `(-2,-1)`

Explanation

To solve the problem step-by-step, we need to find the point \( P \) on the hyperbola defined by \( xy = 2 \) such that the normal drawn from \( P \) to the ellipse defined by \( \frac{x^2}{6} + \frac{y^2}{3} = 1 \) is perpendicular to the line \( x + y = 8 \).

### Step 1: Define the Point \( P \)
Let the coordinates of point \( P \) be \( (h, \frac{2}{h}) \) since \( P \) lies on the curve \( xy = 2 \).

**Hint:** Use the relationship \( xy = 2 \) to express one variable in terms of the other.

### Step 2: Find the Slope of the Line \( x + y = 8 \)
The line \( x + y = 8 \) can be rewritten in slope-intercept form as \( y = -x + 8 \). The slope \( m_1 \) of this line is \( -1 \).

**Hint:** Convert the line equation to slope-intercept form to easily identify the slope.

### Step 3: Determine the Slope of the Normal
Since the normal to the ellipse is perpendicular to the line \( x + y = 8 \), the slope \( m_2 \) of the normal must satisfy:
\[ m_1 \cdot m_2 = -1 \implies -1 \cdot m_2 = -1 \implies m_2 = 1 \]

**Hint:** Use the property of perpendicular lines where the product of their slopes equals -1.

### Step 4: Write the Equation of the Normal
The equation of the normal to the ellipse at a point \( (x_0, y_0) \) can be expressed as:
\[ y - y_0 = m_2 (x - x_0) \]
Substituting \( m_2 = 1 \) gives:
\[ y - y_0 = (x - x_0) \]
Rearranging, we have:
\[ y = x - x_0 + y_0 \]

**Hint:** Remember that the slope of the normal is the negative reciprocal of the slope of the tangent.

### Step 5: Find the Coordinates of the Point on the Ellipse
For the ellipse \( \frac{x^2}{6} + \frac{y^2}{3} = 1 \), the coordinates of the point on the ellipse can be expressed in parametric form as:
\[ (x_0, y_0) = (\sqrt{6} \cos \theta, \sqrt{3} \sin \theta) \]

**Hint:** Use parametric equations to express points on the ellipse.

### Step 6: Set Up the Equation for the Normal
The slope of the normal at the point \( (x_0, y_0) \) is given by:
\[ \text{slope of normal} = -\frac{\sqrt{6} \sec \theta}{-\sqrt{3}} = \frac{\sqrt{6}}{\sqrt{3}} \sec \theta = \sqrt{2} \sec \theta \]
Setting this equal to \( 1 \) gives:
\[ \sqrt{2} \sec \theta = 1 \implies \sec \theta = \frac{1}{\sqrt{2}} \implies \cos \theta = \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4} \]

**Hint:** Use trigonometric identities to relate the slopes.

### Step 7: Substitute \( \theta \) Back to Find \( P \)
Substituting \( \theta = \frac{\pi}{4} \) into the parametric equations gives:
\[ x_0 = \sqrt{6} \cdot \frac{1}{\sqrt{2}} = \sqrt{3}, \quad y_0 = \sqrt{3} \cdot \frac{1}{\sqrt{2}} = \frac{3}{2} \]
Now, we need to relate this back to \( P \):
\[ h = \sqrt{3}, \quad \frac{2}{h} = \frac{2}{\sqrt{3}} \]

**Hint:** Use the values of \( x_0 \) and \( y_0 \) to find \( h \) and subsequently the coordinates of \( P \).

### Step 8: Solve for \( h \)
We can find \( h \) by solving the equation:
\[ h - \frac{2}{h} = 1 \implies h^2 - h - 2 = 0 \]
Factoring gives:
\[ (h - 2)(h + 1) = 0 \implies h = 2 \text{ or } h = -1 \]

**Hint:** Use the quadratic formula or factoring to solve for \( h \).

### Step 9: Find the Corresponding Points \( P \)
For \( h = 2 \):
\[ P_1 = (2, 1) \]
For \( h = -1 \):
\[ P_2 = (-1, -2) \]

**Hint:** Substitute the values of \( h \) back into the expression for \( P \).

### Final Answer
The points \( P \) are \( (2, 1) \) and \( (-1, -2) \). The valid point \( P \) that satisfies all conditions is:
\[ \boxed{(2, 1)} \]