Suppose the ellipse `x^2/2+y^2=1` and the ellipse `x^2/5+y^2/a^2=1` where `a^2=b^2-8b+13` intersect in four distinct points, then number of integral values of b is

Answer

To solve the problem, we need to determine the number of integral values of \( b \) such that the ellipses \( \frac{x^2}{2} + y^2 = 1 \) and \( \frac{x^2}{5} + \frac{y^2}{a^2} = 1 \) intersect at four distinct points. We know that \( a^2 = b^2 - 8b + 13 \).

### Step 1: Analyze the first ellipse
The first ellipse is given by:
\[ \frac{x^2}{2} + y^2 = 1 \]
This can be rewritten as:
\[ y^2 = 1 - \frac{x^2}{2} \]
This implies that the maximum value of \( y^2 \) occurs when \( x = 0 \), giving \( y^2 = 1 \).

### Step 2: Analyze the second ellipse
The second ellipse is given by:
\[ \frac{x^2}{5} + \frac{y^2}{a^2} = 1 \]
Rearranging gives:
\[ y^2 = a^2 \left(1 - \frac{x^2}{5}\right) \]
The maximum value of \( y^2 \) occurs when \( x = 0 \), giving \( y^2 = a^2 \).

### Step 3: Set up the intersection condition
For the ellipses to intersect at four distinct points, the maximum \( y^2 \) values from both ellipses must satisfy:
\[ a^2 > 1 \]
Substituting \( a^2 = b^2 - 8b + 13 \):
\[ b^2 - 8b + 13 > 1 \]
This simplifies to:
\[ b^2 - 8b + 12 > 0 \]

### Step 4: Factor the quadratic inequality
Factoring the quadratic:
\[ (b - 6)(b - 2) > 0 \]
The critical points are \( b = 2 \) and \( b = 6 \). The solution to this inequality is:
\[ b < 2 \quad \text{or} \quad b > 6 \]

### Step 5: Determine integral values of \( b \)
Now we need to find the integral values of \( b \) that satisfy this condition:
- For \( b < 2 \): The integral values are \( \ldots, -1, 0, 1 \) (which are 3 values).
- For \( b > 6 \): The integral values are \( 7, 8, 9, \ldots \) (which are infinite).

### Step 6: Combine the results
Since we are only interested in the integral values of \( b \) that are less than 2 and greater than 6, we can count:
- Values less than 2: \( -1, 0, 1 \) (3 values)
- Values greater than 6: \( 7, 8, 9, \ldots \) (infinitely many)

Thus, the total number of integral values of \( b \) is:
\[ \text{Total integral values of } b = 3 + \text{(infinitely many)} = \infty \]

### Conclusion
The number of integral values of \( b \) is infinite.