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Step-by-Step Solution
1. Write the Equation of the Hyperbola
The given hyperbola is
$$4y^2 = x^2 + 1.$$
Rearranging, we get
$$-\,x^2 + 4y^2 = 1.$$
We can also write this in a standard hyperbola form (though not strictly necessary here) as
$$-\frac{x^2}{1^2} + \frac{y^2}{\left(\frac{1}{2}\right)^2} = 1.$$
2. General Form of a Tangent
For the curve \(4y^2 = x^2 + 1\), let \(\bigl(x_1, y_1\bigr)\) be a point on the hyperbola. We want the equation of the tangent at this point.
Differentiate \(4y^2 = x^2 + 1\) implicitly:
$$
8y\,\frac{dy}{dx} = 2x
\quad \Longrightarrow \quad
\frac{dy}{dx} = \frac{x}{4y}.
$$
Thus, the slope \(m\) of the tangent at \(\bigl(x_1, y_1\bigr)\) is
$$
m = \frac{x_1}{4\,y_1}.
$$
The equation of this tangent line in slope-intercept form is
$$
y = mx + c
\quad \Longrightarrow \quad
y = \frac{x_1}{4\,y_1}\,x + c.
$$
Because the line passes through \(\bigl(x_1, y_1\bigr)\), substitute \(x = x_1,\, y = y_1\) to find \(c\).
$$
y_1 = \frac{x_1}{4\,y_1}\,x_1 + c
\quad \Longrightarrow \quad
c = y_1 - \frac{x_1^2}{4\,y_1}.
$$
3. Tangent in an Alternative Form
Using the condition \(4y_1^2 = x_1^2 + 1\) from the hyperbola, one gets
$$4y_1^2 - x_1^2 = 1,$$
so
$$c = y_1 - \frac{x_1^2}{4y_1} = \frac{4y_1^2 - x_1^2}{4y_1} = \frac{1}{4y_1}.$$
Hence, the tangent line becomes
$$y = \frac{x_1}{4y_1} \, x + \frac{1}{4y_1}.$$
Rearranging,
$$4\,y_1\,y = x_1\,x + 1.$$
4. Intersection with the Coordinate Axes
To find the points where this tangent meets the axes:
x-axis: Set \(y=0\) in \(4\,y_1\,y = x_1\,x + 1\):
$$
0 = x_1\,x + 1 \quad \Longrightarrow \quad x = -\frac{1}{x_1}.
$$
Hence, the point of intersection with the \(x\)-axis is
$$
A\Bigl(-\frac{1}{x_1},\,0\Bigr).
$$
y-axis: Set \(x=0\) in \(4\,y_1\,y = x_1\,x + 1\):
$$
4\,y_1\,y = 1 \quad \Longrightarrow \quad y = \frac{1}{4\,y_1}.
$$
Therefore, the point of intersection with the \(y\)-axis is
$$
B\Bigl(0,\,\tfrac{1}{4\,y_1}\Bigr).
$$
5. Coordinates of the Midpoint (h, k)
The midpoint \((h, k)\) of segment \(AB\) where
\(A\Bigl(-\frac{1}{x_1},0\Bigr)\) and \(B\Bigl(0,\frac{1}{4y_1}\Bigr)\) is found by
\[
h = \frac{-\frac{1}{x_1} + 0}{2} = -\frac{1}{2\,x_1},
\quad
k = \frac{0 + \frac{1}{4y_1}}{2} = \frac{1}{8\,y_1}.
\]
Hence, from \(h = -\frac{1}{2\,x_1}\), we get
$$x_1 = -\frac{1}{2h}.$$
Similarly, from \(k = \frac{1}{8\,y_1}\), we get
$$y_1 = \frac{1}{8k}.$$
6. Eliminate the Parameters to Find the Locus
Recall that \(\bigl(x_1, y_1\bigr)\) lies on the hyperbola
$$4y_1^2 = x_1^2 + 1.$$
We substitute
$$x_1 = -\frac{1}{2h} \quad \text{and} \quad y_1 = \frac{1}{8k}$$
into this equation:
\[
4 \left(\frac{1}{8k}\right)^2 = \left(-\frac{1}{2h}\right)^2 + 1.
\]
That is
\[
4\cdot \frac{1}{64\,k^2}
= \frac{1}{4\,h^2} + 1
\quad \Longrightarrow \quad
\frac{1}{16\,k^2}
= \frac{1}{4\,h^2} + 1.
\]
Multiply through by \(16k^2\):
\[
1 = 4\,k^2 \cdot \frac{1}{h^2} + 16\,k^2.
\]
Rewriting and rearranging in terms of \(h\) and \(k\):
\[
h^2 = 4\,k^2 + 16\,h^2\,k^2.
\]
7. Final Locus Equation
Replacing \(h\) by \(x\) and \(k\) by \(y\), we get the locus of the midpoint \((x, y)\) as
$$
x^2 - 4\,y^2 - 16\,x^2\,y^2 = 0.
$$
Answer
The required locus of the midpoint of \(AB\) is
$$x^2 - 4y^2 \;-\; 16\,x^2\,y^2 = 0.$$
