A tangent to the hyperbola `x^(2)/4 - y^(2)/1 = 1` meets ellipse `x^(2) + 4y^(2) = 4` in two distinct points .
Then the locus of midpoint of this chord is

Options

A. `(x^(2) + 4y^(2))^(2) = 4 (x^(2) - 4y^(2))`
B. `(x^(2) - 4y^(2))^(2) = 4 (x^(2) + 4y^(2)) `
C. `(x^(2) - 4y^(2))^(2) = 4(x^(2) + y^(2))`
D. none of these

Answer

To find the locus of the midpoint of the chord formed by the intersection of a tangent to the hyperbola \( \frac{x^2}{4} - \frac{y^2}{1} = 1 \) and the ellipse \( x^2 + 4y^2 = 4 \), we can follow these steps:

### Step 1: Identify the equations of the hyperbola and ellipse
The hyperbola is given by:
\[ \frac{x^2}{4} - \frac{y^2}{1} = 1 \]
This can be rewritten in standard form as:
\[ x^2 - 4y^2 = 4 \]

The ellipse is given by:
\[ x^2 + 4y^2 = 4 \]

### Step 2: Write the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola at point \( (x_0, y_0) \) can be expressed in terms of the slope \( m \):
\[ y = mx \pm \sqrt{4m^2 - 1} \]

### Step 3: Find the midpoint of the chord
Let the midpoint of the chord be \( (h, k) \). The equation of the chord can be expressed using the midpoint formula:
\[ hx + 4ky = 4 \]

### Step 4: Use the condition for the chord of contact
The condition for the chord of contact for the ellipse is given by:
\[ hx + 4ky = 4 \]
This means that the point \( (h, k) \) lies on the ellipse.

### Step 5: Substitute the midpoint into the ellipse equation
Substituting \( (h, k) \) into the ellipse equation:
\[ h^2 + 4k^2 = 4 \]

### Step 6: Relate the tangent slope to the midpoint
From the tangent equation, we know:
\[ m = \frac{y - k}{x - h} \]
We can express the slope \( m \) in terms of \( h \) and \( k \).

### Step 7: Set up the equation for the locus
Using the relationship between the slope and the coordinates, we can derive an equation. The condition for the tangent to intersect the ellipse in two distinct points leads to:
\[ 4h^2 - 16k^2 = (h^2 + 4k^2)^2 \]

### Step 8: Rearranging the equation
Rearranging gives us:
\[ 4h^2 - 16k^2 = h^4 + 8h^2k^2 + 16k^4 \]

### Step 9: Identify the locus
By replacing \( h \) with \( x \) and \( k \) with \( y \), we can express the locus of the midpoint:
\[ 4x^2 - 16y^2 = (x^2 + 4y^2)^2 \]

### Final Answer
The locus of the midpoint of the chord is given by:
\[ 4x^2 - 16y^2 = (x^2 + 4y^2)^2 \]