If the line `2px+ysqrt(5-6p^(2))=1, p in [-sqrt(5)/(6),sqrt(5)/(6)]`, always touches the standard ellipse. Then find the eccentricity of the standard ellipse.

Explanation

To solve the problem step by step, we need to analyze the given line and its relationship with the standard ellipse.

### Step 1: Understand the given line equation
The line is given by:
\[ 2px + y \sqrt{5 - 6p^2} = 1 \]
where \( p \) is in the interval \( \left[-\frac{\sqrt{5}}{6}, \frac{\sqrt{5}}{6}\right] \).

### Step 2: Rearranging the line equation
We can express \( y \) in terms of \( x \) and \( p \):
\[ y = \frac{1 - 2px}{\sqrt{5 - 6p^2}} \]

### Step 3: Identify the slope of the line
From the rearranged equation, we can identify the slope \( m \):
\[ m = -\frac{2p}{\sqrt{5 - 6p^2}} \]

### Step 4: Equation of the tangent to the ellipse
For a standard ellipse represented by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
the equation of the tangent line with slope \( m \) is given by:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]

### Step 5: Setting the equations equal
Since the line touches the ellipse, we equate the two expressions for \( y \):
\[ \frac{1 - 2px}{\sqrt{5 - 6p^2}} = mx + \sqrt{a^2 m^2 + b^2} \]

### Step 6: Substitute the value of \( m \)
Substituting \( m = -\frac{2p}{\sqrt{5 - 6p^2}} \) into the tangent equation:
\[ \frac{1 - 2px}{\sqrt{5 - 6p^2}} = -\frac{2p}{\sqrt{5 - 6p^2}}x + \sqrt{a^2 \left(-\frac{2p}{\sqrt{5 - 6p^2}}\right)^2 + b^2} \]

### Step 7: Simplifying the equation
Cross-multiplying and simplifying gives:
\[ 1 - 2px = -2px + \sqrt{a^2 \frac{4p^2}{5 - 6p^2} + b^2} \cdot \sqrt{5 - 6p^2} \]

### Step 8: Collect terms
Rearranging leads to:
\[ 1 = \sqrt{a^2 \frac{4p^2}{5 - 6p^2} + b^2} \cdot \sqrt{5 - 6p^2} \]

### Step 9: Square both sides
Squaring both sides results in:
\[ 1 = a^2 \frac{4p^2}{5 - 6p^2} + b^2 \]

### Step 10: Analyze the conditions
Since this must hold for all \( p \) in the given range, we can analyze the coefficients of \( p^2 \) and the constant terms separately.

### Step 11: Set up equations
From the equation:
1. \( 4a^2 - 6b^2 = 0 \)
2. \( 5b^2 - 1 = 0 \)

### Step 12: Solve for \( b^2 \)
From the second equation:
\[ b^2 = \frac{1}{5} \]

### Step 13: Substitute \( b^2 \) into the first equation
Substituting \( b^2 \) into the first equation:
\[ 4a^2 - 6 \cdot \frac{1}{5} = 0 \]
\[ 4a^2 = \frac{6}{5} \]
\[ a^2 = \frac{3}{10} \]

### Step 14: Calculate the eccentricity
The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values:
\[ e = \sqrt{1 - \frac{\frac{1}{5}}{\frac{3}{10}}} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \]

### Final Answer
Thus, the eccentricity of the standard ellipse is:
\[ e = \frac{1}{\sqrt{3}} \]