The product of the slopes of the tangents to the ellipse `2x^(2)+3y^(2)=6` drawn from the point (1, 2) is

Options

A. 1
B. 2
C. `-1`
D. `-2`

Answer

To find the product of the slopes of the tangents to the ellipse \(2x^2 + 3y^2 = 6\) drawn from the point \((1, 2)\), we will follow these steps:

### Step 1: Rewrite the equation of the ellipse in standard form
The given equation of the ellipse is:
\[ 2x^2 + 3y^2 = 6 \]
Dividing the entire equation by 6 gives us:
\[ \frac{x^2}{3} + \frac{y^2}{2} = 1 \]
This indicates that \(a^2 = 3\) and \(b^2 = 2\).

### Step 2: Write the equation of the tangent to the ellipse
The general equation of the tangent to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is given by:
\[ y = mx \pm \sqrt{a^2m^2 + b^2} \]
Substituting \(a^2\) and \(b^2\):
\[ y = mx \pm \sqrt{3m^2 + 2} \]

### Step 3: Substitute the point (1, 2) into the tangent equation
Since the point \((1, 2)\) lies on the tangents, we substitute \(x = 1\) and \(y = 2\):
\[ 2 = m(1) \pm \sqrt{3m^2 + 2} \]
This simplifies to:
\[ 2 = m \pm \sqrt{3m^2 + 2} \]

### Step 4: Rearrange and square both sides
Rearranging gives:
\[ 2 - m = \pm \sqrt{3m^2 + 2} \]
Now, squaring both sides:
\[ (2 - m)^2 = 3m^2 + 2 \]
Expanding the left side:
\[ 4 - 4m + m^2 = 3m^2 + 2 \]
Rearranging gives:
\[ 4 - 2 = 3m^2 - m^2 + 4m \]
This simplifies to:
\[ 2m^2 + 4m + 2 = 0 \]

### Step 5: Simplify the quadratic equation
Dividing the entire equation by 2:
\[ m^2 + 2m + 1 = 0 \]
This factors to:
\[ (m + 1)^2 = 0 \]
Thus, the roots are:
\[ m_1 = -1, \quad m_2 = -1 \]

### Step 6: Calculate the product of the slopes
The product of the slopes \(m_1\) and \(m_2\) is given by:
\[ m_1 \cdot m_2 = (-1) \cdot (-1) = 1 \]

### Final Answer
The product of the slopes of the tangents to the ellipse drawn from the point \((1, 2)\) is:
\[ \boxed{1} \]