P(h, k) is a point in the xy plane such that two perpendicular tangents can be drawn from P to `3x^2 +4y^2=12`. Then `(h^2+k^2)/2` equals

Answer

To solve the problem, we need to find the value of \((h^2 + k^2)/2\) given that two perpendicular tangents can be drawn from the point \(P(h, k)\) to the ellipse defined by the equation \(3x^2 + 4y^2 = 12\).

### Step-by-step Solution:

1. **Rewrite the Ellipse Equation**:
The given ellipse equation is \(3x^2 + 4y^2 = 12\). We can rewrite it in standard form:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]
Here, \(a^2 = 4\) and \(b^2 = 3\).

**Hint**: Identify the semi-major and semi-minor axes from the standard form of the ellipse.

2. **Identify the Condition for Perpendicular Tangents**:
The condition for a point \(P(h, k)\) to have two perpendicular tangents to the ellipse is given by the equation:
\[ h^2 + k^2 = a^2 + b^2 \]
where \(a^2\) and \(b^2\) are the denominators from the standard form of the ellipse.

**Hint**: Remember that the sum of the squares of the semi-major and semi-minor axes gives the locus of points from which perpendicular tangents can be drawn.

3. **Calculate \(a^2 + b^2\)**:
We have:
\[ a^2 = 4 \quad \text{and} \quad b^2 = 3 \]
Therefore:
\[ a^2 + b^2 = 4 + 3 = 7 \]

**Hint**: Sum the values of \(a^2\) and \(b^2\) to find the required value.

4. **Set Up the Equation**:
From the condition for perpendicular tangents, we have:
\[ h^2 + k^2 = 7 \]

**Hint**: This equation directly relates the coordinates of the point \(P\) to the ellipse.

5. **Find \((h^2 + k^2)/2\)**:
We need to find:
\[ \frac{h^2 + k^2}{2} = \frac{7}{2} \]

**Hint**: Divide the sum \(h^2 + k^2\) by 2 to obtain the final result.

### Final Answer:
\[ \frac{h^2 + k^2}{2} = \frac{7}{2} \]