Number of points from which two perpendicular tangents can be drawn to both the ellipses `E_1: x^2/(a^2+2)+y^2/b^2=1` and `E_2:x^2/a^2+y^2/(b^2+1)=1` is

Options

A. 0
B. 1
C. 2
D. infinitely many

Answer

To solve the problem of finding the number of points from which two perpendicular tangents can be drawn to both ellipses \( E_1: \frac{x^2}{a^2 + 2} + \frac{y^2}{b^2} = 1 \) and \( E_2: \frac{x^2}{a^2} + \frac{y^2}{b^2 + 1} = 1 \), we will follow these steps:

### Step 1: Understand the Condition for Perpendicular Tangents
For two tangents to be perpendicular 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 2: Write the Equation of Tangents
For an ellipse given by the equation \( \frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1 \), the equation of the tangent at point \( (x_0, y_0) \) can be expressed as:
\[ y = mx \pm \sqrt{\alpha^2 m^2 + \beta^2} \]

### Step 3: Apply to the First Ellipse \( E_1 \)
For the first ellipse \( E_1: \frac{x^2}{a^2 + 2} + \frac{y^2}{b^2} = 1 \):
- The equation of the tangent can be written as:
\[ y = mx \pm \sqrt{(a^2 + 2)m^2 + b^2} \]

### Step 4: Apply to the Second Ellipse \( E_2 \)
For the second ellipse \( E_2: \frac{x^2}{a^2} + \frac{y^2}{b^2 + 1} = 1 \):
- The equation of the tangent can be written as:
\[ y = mx \pm \sqrt{a^2 m^2 + (b^2 + 1)} \]

### Step 5: Set Up the Condition for Perpendicular Tangents
For both ellipses, we need to find points where the tangents can be perpendicular. This leads to the following equations based on the slopes:
1. From \( E_1 \):
\[ \sqrt{(a^2 + 2)m^2 + b^2} \]
2. From \( E_2 \):
\[ \sqrt{a^2 m^2 + (b^2 + 1)} \]

### Step 6: Equate and Solve
To find the points where perpendicular tangents can be drawn, we need to equate the conditions derived from both ellipses. This leads to a system of equations that we need to solve.

### Step 7: Analyze the Result
After setting up the equations and simplifying, we find that there are no real solutions for \( x \) and \( y \) that satisfy both conditions simultaneously.

### Conclusion
Since there are no real values of \( x \) and \( y \) that satisfy the conditions for both ellipses, we conclude that the number of points from which two perpendicular tangents can be drawn to both ellipses is:
\[ \text{Number of points} = 0 \]