The given differential equation is (a^2 - x^2)p^2 + 2xyp + (b^2 - y^2) = 0.

To find the envelope of this family of straight lines, we need to eliminate the arbitrary constant (p) from the equation.

Let's solve the given equation step by step:

1. Rewrite the equation in the form of a quadratic equation in terms of p:
(a^2 - x^2)p^2 + 2xyp + (b^2 - y^2) = 0

2. Applying the quadratic formula, we have:
p = (-2xy ± √((2xy)^2 - 4(a^2 - x^2)(b^2 - y^2))) / 2(a^2 - x^2)

3. Simplify the equation:
p = -xy ± √(x^2y^2 - (a^2 - x^2)(b^2 - y^2)) / (a^2 - x^2)

4. Consider the equation p = -xy ± √(x^2y^2 - (a^2 - x^2)(b^2 - y^2)) / (a^2 - x^2)

5. We can see that p is a function of x and y, and it depends on the values of a and b.

6. To find the envelope, we need to eliminate p from the equation. This can be done by taking the derivative of p with respect to x and y and setting it equal to zero.

7. Taking the derivative of p with respect to x:
dp/dx = -y - [(2x)(-2(a^2 - x^2)(-y))] / (a^2 - x^2)^2

8. Taking the derivative of p with respect to y:
dp/dy = -x - [(2y)(-2(a^2 - x^2)(b^2 - y^2))] / (a^2 - x^2)^2

9. Setting both derivatives equal to zero, we have:
-y - [(2x)(-2(a^2 - x^2)(-y))] / (a^2 - x^2)^2 = 0
-x - [(2y)(-2(a^2 - x^2)(b^2 - y^2))] / (a^2 - x^2)^2 = 0

10. Simplifying the equations, we get:
y(a^2 - x^2) = 0
x(b^2 - y^2) = 0

11. Solving for y, we have:
y = 0 or y = ±a

12. Solving for x, we have:
x = 0 or x = ±b

13. Therefore, the envelope of the family of straight lines is given by the equations y = 0, y = ±a, x = 0, and x = ±b.

14. The envelope represents an ellipse, where the major axis is along the y-axis with endpoints (0, ±a), and the minor axis is along the x-axis with endpoints (±b, 0).

Hence, the correct answer is option C, ellipse.