If a line `3 px + 2ysqrt(1-p^2) =1` touches a fixed ellipse E for all p `in` [- 1, 1], then equation of a directrix of E the ellipse is:

Options

A. `x=(3sqrt5)/10`
B. `y=(3sqrt5)/10`
C. `x=sqrt5/3`
D. `y=sqrt5/3`

Answer

To solve the problem step by step, we need to analyze the given line and its relationship with the ellipse. The line is given as:

\[ 3px + 2y\sqrt{1 - p^2} = 1 \]

### Step 1: Identify the form of the ellipse
We start by assuming the standard form of the ellipse, which can be expressed as:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

Here, \(a\) and \(b\) are the semi-major and semi-minor axes respectively.

### Step 2: Understand the condition of tangency
The line touches the ellipse for all values of \(p\) in the interval \([-1, 1]\). This implies that the line is tangent to the ellipse. For a line to be tangent to an ellipse, the discriminant of the quadratic equation formed must be zero.

### Step 3: Rearranging the line equation
We can rearrange the line equation to express \(y\) in terms of \(x\):

\[ 2y\sqrt{1 - p^2} = 1 - 3px \]
\[ y = \frac{1 - 3px}{2\sqrt{1 - p^2}} \]

### Step 4: Substitute into the ellipse equation
Substituting \(y\) into the ellipse equation gives us a quadratic in \(p\):

\[ \frac{x^2}{a^2} + \frac{\left(\frac{1 - 3px}{2\sqrt{1 - p^2}}\right)^2}{b^2} = 1 \]

### Step 5: Simplifying the equation
After substituting, we need to simplify the equation and collect terms to form a quadratic in \(p\):

1. Expand the \(y\) term.
2. Combine like terms to form a quadratic equation in \(p\).

### Step 6: Set the discriminant to zero
For the line to be tangent to the ellipse, the discriminant of the quadratic must be zero:

\[ D = B^2 - 4AC = 0 \]

Where \(A\), \(B\), and \(C\) are the coefficients of the quadratic equation in \(p\).

### Step 7: Find the values of \(a\), \(b\), and eccentricity \(e\)
From the quadratic formed, we can find the values of \(a\) and \(b\) by comparing coefficients. The eccentricity \(e\) of the ellipse can be calculated using the formula:

\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]

### Step 8: Directrix of the ellipse
The equations of the directrices for the ellipse are given by:

1. \(x = \pm \frac{a}{e}\) (for horizontal ellipses)
2. \(y = \pm \frac{b}{e}\) (for vertical ellipses)

Since we determined that the major axis is along the y-axis, we will use the second form.

### Step 9: Substitute the values to find the directrix
Substituting the values of \(b\) and \(e\) into the directrix equation:

\[ y = \pm \frac{b}{e} \]

### Final Result
After performing all the calculations, we find the equations of the directrices of the ellipse.

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