Let `E_1` be the ellipse `x^2/(a^2+2)+y^2/b^2=1` and `E_2` be the ellipse `x^2/a^2+y^2/(b^2+1)=1`. The number of points from which two perpendicular tangents can be drawn to each of `E_1` and `E_2` is

Answer

To solve the problem of finding the number of points from which two perpendicular tangents can be drawn to the ellipses \( E_1 \) and \( E_2 \), we will follow these steps:

### Step 1: Understand the equations of the ellipses
The equations of the ellipses are given as:
1. \( E_1: \frac{x^2}{a^2 + 2} + \frac{y^2}{b^2} = 1 \)
2. \( E_2: \frac{x^2}{a^2} + \frac{y^2}{b^2 + 1} = 1 \)

### Step 2: Find the condition for perpendicular tangents
For a point \( (x_0, y_0) \) to have two perpendicular tangents to an ellipse, the slopes of the tangents must satisfy the condition \( m_1 \cdot m_2 = -1 \), where \( m_1 \) and \( m_2 \) are the slopes of the tangents.

### Step 3: Derive the equation for \( E_1 \)
For the ellipse \( E_1 \), the equation of the tangent at point \( (x_0, y_0) \) can be written as:
\[ y - y_0 = m(x - x_0) \quad \text{(where \( m \) is the slope)} \]
Substituting this into the ellipse equation leads to:
\[ \frac{x_0^2}{a^2 + 2} + \frac{(mx + y_0 - mx_0)^2}{b^2} = 1 \]
This will yield a quadratic equation in \( m \).

### Step 4: Derive the equation for \( E_2 \)
Similarly, for the ellipse \( E_2 \), the equation of the tangent can be derived:
\[ y - y_0 = m(x - x_0) \]
Substituting this into the ellipse equation leads to:
\[ \frac{x_0^2}{a^2} + \frac{(mx + y_0 - mx_0)^2}{b^2 + 1} = 1 \]
This will also yield a quadratic equation in \( m \).

### Step 5: Set up the equations for perpendicular tangents
For both ellipses, we need to find the conditions under which the product of the slopes \( m_1 \cdot m_2 = -1 \). This leads to two equations that must be satisfied simultaneously.

### Step 6: Solve the equations
By equating the derived conditions from both ellipses, we can find the relationship between \( x \) and \( y \). After simplifying the equations, we arrive at:
\[ x^2 + y^2 = a^2 + b^2 + 2 \]
and
\[ x^2 + y^2 = a^2 + b^2 + 1 \]

### Step 7: Analyze the results
Subtracting the two equations gives:
\[ 0 = 1 \]
This indicates that there are no real solutions for \( x \) and \( y \) that satisfy both conditions simultaneously.

### Conclusion
Thus, the number of points from which two perpendicular tangents can be drawn to both ellipses \( E_1 \) and \( E_2 \) is **0**.