An ellipse has eccentricity `1/2` and one focus at the point `P(1/2,1)`. Its one directrix is the common tangent nearer to the point the P to the hyperbola `x^2-y^2=1` and the circle `x^2+y^2=1`.The equation of the ellipse , in the standard form is-

Options

A. `9(x-1/3)^2 +12(y-1)^2 = 1`
B. `8(x-1/3)^2 +12(y-1)^2 = 1`
C. `9(x-1/3)^2 +8(y-1)^2 = 1`
D. none of these

Answer

To find the equation of the ellipse with the given properties, we will follow these steps:

### Step 1: Identify the focus and eccentricity
The focus of the ellipse is given as \( P\left(\frac{1}{2}, 1\right) \) and the eccentricity \( e = \frac{1}{2} \).

### Step 2: Determine the directrix
The directrix of the ellipse is the common tangent nearer to the point \( P \) to the hyperbola \( x^2 - y^2 = 1 \) and the circle \( x^2 + y^2 = 1 \). The equation of the directrix can be found by determining the tangent lines to both curves.

The hyperbola \( x^2 - y^2 = 1 \) has a tangent line at the point where it intersects the x-axis, which can be approximated as \( x = 1 \) (the nearest point to \( P \)). The circle \( x^2 + y^2 = 1 \) has a tangent line at the point \( (1,0) \), which is also \( x = 1 \).

Thus, the directrix is \( x = 1 \).

### Step 3: Use the relationship between focus, directrix, and eccentricity
The relationship between the focus \( (x_0, y_0) \), the directrix \( x = d \), and the eccentricity \( e \) is given by:

\[ \frac{\sqrt{(x_0 - d)^2 + (y_0 - 0)^2}}{e} = |x_0 - d| \]

Substituting the values we have:

\[ \frac{\sqrt{\left(\frac{1}{2} - 1\right)^2 + (1 - 0)^2}}{\frac{1}{2}} = \left|\frac{1}{2} - 1\right| \]

Calculating the left side:

\[ \frac{\sqrt{\left(-\frac{1}{2}\right)^2 + 1^2}}{\frac{1}{2}} = \frac{\sqrt{\frac{1}{4} + 1}}{\frac{1}{2}} = \frac{\sqrt{\frac{5}{4}}}{\frac{1}{2}} = \frac{\sqrt{5}}{2} \cdot 2 = \sqrt{5} \]

Calculating the right side:

\[ \left|\frac{1}{2} - 1\right| = \left|-\frac{1}{2}\right| = \frac{1}{2} \]

### Step 4: Solve for \( a \) and \( b \)
From the relationship \( |x_0 - d| = ae \), we have:

\[ \sqrt{5} = a \cdot \frac{1}{2} \]

Thus,

\[ a = 2\sqrt{5} \]

Now, using the relationship \( e = \frac{c}{a} \) where \( c = \sqrt{a^2 - b^2} \):

\[ \frac{1}{2} = \frac{c}{2\sqrt{5}} \implies c = \sqrt{5} \]

Now, we know:

\[ c^2 = a^2 - b^2 \implies 5 = (2\sqrt{5})^2 - b^2 \implies 5 = 20 - b^2 \implies b^2 = 15 \]

### Step 5: Write the standard form of the ellipse
The standard form of the ellipse centered at \( (0, 1) \) is:

\[ \frac{(x - 0)^2}{(2\sqrt{5})^2} + \frac{(y - 1)^2}{15} = 1 \]

This simplifies to:

\[ \frac{x^2}{20} + \frac{(y - 1)^2}{15} = 1 \]

### Final Answer
The equation of the ellipse in standard form is:

\[ \frac{x^2}{20} + \frac{(y - 1)^2}{15} = 1 \]