If the tangent drawn to the hyperbola `4y^2=x^2+1` intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid-point of AB is:

Options

A. `4x^2-y^2-16x^2y^2=0`
B. `4x^2-y^2+16x^2y^2=0`
C. `x^2-4y^2+16x^2y^2=0`
D. `x^2-4y^2-16x^2y^2=0`

Answer

To find the locus of the midpoint of the points where the tangent to the hyperbola \(4y^2 = x^2 + 1\) intersects the coordinate axes, we can follow these steps:

### Step 1: Rewrite the Hyperbola in Standard Form
The given hyperbola is:
\[ 4y^2 = x^2 + 1 \]
Rearranging gives:
\[ -x^2 + 4y^2 = 1 \]
This can be rewritten as:
\[ \frac{y^2}{\frac{1}{4}} - \frac{x^2}{1} = 1 \]
This shows that it is a hyperbola centered at the origin with \(a^2 = 1\) and \(b^2 = \frac{1}{4}\).

### Step 2: Find the Equation of the Tangent
Let the point of tangency on the hyperbola be \((x_1, y_1)\). The equation of the tangent at this point can be derived from the hyperbola's equation:
\[ \frac{yy_1}{\frac{1}{4}} - \frac{xx_1}{1} = 1 \]
This simplifies to:
\[ 4yy_1 - xx_1 = 1 \]

### Step 3: Determine Intersections with the Axes
To find the intersection with the x-axis, set \(y = 0\):
\[ 4(0)y_1 - xx_1 = 1 \implies -xx_1 = 1 \implies x = -\frac{1}{x_1} \]
Thus, the intersection point on the x-axis is \(\left(-\frac{1}{x_1}, 0\right)\).

To find the intersection with the y-axis, set \(x = 0\):
\[ 4yy_1 - 0 = 1 \implies 4yy_1 = 1 \implies y = \frac{1}{4y_1} \]
Thus, the intersection point on the y-axis is \(\left(0, \frac{1}{4y_1}\right)\).

### Step 4: Find the Midpoint of Points A and B
Let \(A = \left(-\frac{1}{x_1}, 0\right)\) and \(B = \left(0, \frac{1}{4y_1}\right)\). The midpoint \(M\) of \(AB\) is given by:
\[ M = \left(\frac{-\frac{1}{x_1} + 0}{2}, \frac{0 + \frac{1}{4y_1}}{2}\right) = \left(-\frac{1}{2x_1}, \frac{1}{8y_1}\right) \]

### Step 5: Express \(x_1\) and \(y_1\) in Terms of \(h\) and \(k\)
Let \(M = (h, k)\). Then:
\[ h = -\frac{1}{2x_1} \implies x_1 = -\frac{1}{2h} \]
\[ k = \frac{1}{8y_1} \implies y_1 = \frac{1}{8k} \]

### Step 6: Substitute into the Hyperbola Equation
Using the hyperbola equation:
\[ 4y_1^2 = x_1^2 + 1 \]
Substituting \(y_1\) and \(x_1\):
\[ 4\left(\frac{1}{8k}\right)^2 = \left(-\frac{1}{2h}\right)^2 + 1 \]
This simplifies to:
\[ \frac{4}{64k^2} = \frac{1}{4h^2} + 1 \]
\[ \frac{1}{16k^2} = \frac{1}{4h^2} + 1 \]

### Step 7: Rearranging the Equation
Multiplying through by \(16k^2\):
\[ 1 = 4k^2 \cdot \frac{1}{4h^2} + 16k^2 \]
\[ 1 = \frac{4k^2}{4h^2} + 16k^2 \]
\[ 1 = \frac{k^2}{h^2} + 16k^2 \]
Now, rearranging gives:
\[ h^2 = 4k^2 + 16k^2 \]
\[ h^2 = 20k^2 \]

### Final Step: The Locus Equation
This can be rewritten as:
\[ \frac{h^2}{20} - k^2 = 0 \]
Thus, the locus of the midpoint of \(AB\) is:
\[ \frac{x^2}{20} - y^2 = 0 \]