For all real p, the line `2px+ysqrt(1-p^(2))=1` touches a fixed ellipse whose axex are the coordinate axes
The foci of the ellipse are

Options

A. `(pmsqrt3/2,0)`
B. `(0,pmsqrt3/2)`
C. `(pmsqrt3,0)`
D. `(0,pmsqrt3)`

Answer

To solve the problem, we need to find the foci of the ellipse given that the line \(2px + y\sqrt{1 - p^2} = 1\) touches it for all real \(p\). The ellipse has its axes along the coordinate axes.

### Step-by-Step Solution:

1. **Identify the General Equation of the Ellipse**:
The general equation of an ellipse with axes along the coordinate axes is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

2. **Rewrite the Given Line Equation**:
The given line is:
\[ 2px + y\sqrt{1 - p^2} = 1 \]
We can rearrange this to express \(y\) in terms of \(x\):
\[ y\sqrt{1 - p^2} = 1 - 2px \quad \Rightarrow \quad y = \frac{1 - 2px}{\sqrt{1 - p^2}} \]

3. **Identify the Slope and Intercept**:
From the rearranged line equation, we can identify the slope \(m\) and the y-intercept:
\[ m = -\frac{2p}{\sqrt{1 - p^2}}, \quad \text{and the y-intercept is } \frac{1}{\sqrt{1 - p^2}}. \]

4. **Condition for Tangency**:
For the line to touch the ellipse, the condition is given by:
\[ y = mx \pm \sqrt{a^2m^2 + b^2} \]
By comparison, we have:
\[ \sqrt{1 - p^2} = \sqrt{a^2m^2 + b^2} \]

5. **Substituting the Slope**:
Substitute \(m\) into the equation:
\[ 1 - p^2 = a^2\left(-\frac{2p}{\sqrt{1 - p^2}}\right)^2 + b^2 \]
Simplifying this gives:
\[ 1 - p^2 = \frac{4a^2p^2}{1 - p^2} + b^2 \]

6. **Rearranging the Equation**:
Multiply through by \(1 - p^2\):
\[ (1 - p^2)(1 - p^2) = 4a^2p^2 + b^2(1 - p^2) \]
Expanding and rearranging leads to:
\[ 1 - 2p^2 + p^4 = 4a^2p^2 + b^2 - b^2p^2 \]

7. **Collecting Terms**:
Collecting terms involving \(p^2\):
\[ p^4 + (b^2 - 4a^2 - 2)p^2 + (1 - b^2) = 0 \]

8. **Condition for Real \(p\)**:
For this quadratic in \(p^2\) to have real solutions, the discriminant must be non-negative:
\[ (b^2 - 4a^2 - 2)^2 - 4(1 - b^2) \geq 0 \]

9. **Solving for \(a\) and \(b\)**:
Setting \(1 - b^2 = 0\) gives \(b^2 = 1\). Substituting this back into the equation gives:
\[ 4a^2 - 1 = 0 \quad \Rightarrow \quad a^2 = \frac{1}{4} \quad \Rightarrow \quad a = \frac{1}{2} \]

10. **Finding the Foci**:
The foci of the ellipse are given by \((\pm ae, 0)\) where \(e = \sqrt{1 - \frac{b^2}{a^2}}\):
\[ e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]
Thus, the foci are:
\[ \left(\pm \frac{1}{2} \cdot \frac{\sqrt{3}}{2}, 0\right) = \left(\pm \frac{\sqrt{3}}{4}, 0\right) \]

### Final Answer:
The foci of the ellipse are \(\left(\frac{\sqrt{3}}{4}, 0\right)\) and \(\left(-\frac{\sqrt{3}}{4}, 0\right)\).