The line`2px+ysqrt(1-p^(2))=1(abs(p)lt1)` for different values of p, touches a fixed ellipse whose exes are the coordinate axes. Q. The eccentricity of the ellipse is

Options

A. `1/sqrt5`
B. `1/sqrt3`
C. `sqrt3/2`
D. `2/sqrt5`

Answer

To find the eccentricity of the ellipse whose axes are the coordinate axes and which is touched by the line given by the equation \( 2px + y\sqrt{1 - p^2} = 1 \) for different values of \( p \) (where \( |p| < 1 \)), we can follow these steps:

### Step 1: Write the equation of the ellipse
Since the axes of the ellipse are the coordinate axes, the standard form of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

### Step 2: Rewrite the line equation in slope-intercept form
The given line equation is:
\[ 2px + y\sqrt{1 - p^2} = 1 \]
Rearranging this gives:
\[ y\sqrt{1 - p^2} = 1 - 2px \]
\[ y = \frac{1}{\sqrt{1 - p^2}} - \frac{2p}{\sqrt{1 - p^2}} x \]
Here, the slope \( m \) of the line is:
\[ m = -\frac{2p}{\sqrt{1 - p^2}} \]

### Step 3: Use the condition for tangency
For the line to be a tangent to the ellipse, we can use the formula for the tangent line to an ellipse:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
Comparing this with our line equation, we can set:
\[ \sqrt{a^2 m^2 + b^2} = \frac{1}{\sqrt{1 - p^2}} \]

### Step 4: Substitute \( m \) into the tangency condition
Substituting \( m \) into the equation gives:
\[ \sqrt{a^2 \left(-\frac{2p}{\sqrt{1 - p^2}}\right)^2 + b^2} = \frac{1}{\sqrt{1 - p^2}} \]
Squaring both sides results in:
\[ a^2 \frac{4p^2}{1 - p^2} + b^2 = \frac{1}{1 - p^2} \]

### Step 5: Rearranging and simplifying
Multiplying through by \( 1 - p^2 \):
\[ 4a^2 p^2 + b^2(1 - p^2) = 1 \]
This simplifies to:
\[ 4a^2 p^2 + b^2 - b^2 p^2 = 1 \]
Rearranging gives:
\[ (4a^2 - b^2)p^2 + b^2 - 1 = 0 \]

### Step 6: For the quadratic in \( p^2 \) to have real roots
The discriminant of this quadratic must be zero for the line to be tangent to the ellipse:
\[ D = 0 \implies (b^2 - 1)^2 - 4(4a^2 - b^2)(0) = 0 \]
This implies:
\[ b^2 - 1 = 0 \implies b^2 = 1 \]

### Step 7: Substitute \( b^2 \) back into the equation
Substituting \( b^2 = 1 \) into the previous equation gives:
\[ 4a^2 - 1 = 0 \implies 4a^2 = 1 \implies a^2 = \frac{1}{4} \]

### Step 8: Calculate the eccentricity
The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{\frac{1}{4}}{1}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Conclusion
The eccentricity of the ellipse is:
\[ \frac{\sqrt{3}}{2} \]