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 segment formed by the intersection of the tangent to the hyperbola \(4y^2 = x^2 + 1\) with 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 expressed in standard form as:
\[ \frac{x^2}{1} - \frac{y^2}{\frac{1}{4}} = -1 \]
Thus, we identify \(a^2 = 1\) and \(b^2 = \frac{1}{4}\), leading to \(a = 1\) and \(b = \frac{1}{2}\).

### Step 2: Parametric Equation of the Tangent
The parametric equation of the tangent to the hyperbola at an angle \(\theta\) is given by:
\[ \frac{x \tan \theta}{a} - \frac{y \sec \theta}{b} = 1 \]
Substituting \(a\) and \(b\):
\[ \frac{x \tan \theta}{1} - \frac{y \sec \theta}{\frac{1}{2}} = 1 \]
This simplifies to:
\[ x \tan \theta - 2y \sec \theta = 1 \]

### Step 3: Find Intercepts on the Axes
To find the x-intercept (where \(y = 0\)):
\[ x \tan \theta = 1 \implies x = \frac{1}{\tan \theta} \]
To find the y-intercept (where \(x = 0\)):
\[ -2y \sec \theta = 1 \implies y = -\frac{1}{2 \sec \theta} = -\frac{1}{2} \cos \theta \]

### Step 4: Midpoint of Segment AB
Let the coordinates of points A and B be:
- A: \(\left(\frac{1}{\tan \theta}, 0\right)\)
- B: \(\left(0, -\frac{1}{2} \cos \theta\right)\)

The midpoint \(M(h, k)\) of segment AB is given by:
\[ h = \frac{\frac{1}{\tan \theta} + 0}{2} = \frac{1}{2 \tan \theta} \]
\[ k = \frac{0 - \frac{1}{2} \cos \theta}{2} = -\frac{1}{4} \cos \theta \]

### Step 5: Express \(\tan \theta\) and \(\sec \theta\) in terms of \(h\) and \(k\)
From \(h = \frac{1}{2 \tan \theta}\), we have:
\[ \tan \theta = \frac{1}{2h} \]
From \(k = -\frac{1}{4} \cos \theta\), we have:
\[ \cos \theta = -4k \]
Using the identity \(\sec^2 \theta - \tan^2 \theta = 1\):
\[ \sec^2 \theta = \frac{1}{\cos^2 \theta} = \frac{1}{(-4k)^2} = \frac{1}{16k^2} \]
Thus:
\[ \sec^2 \theta - \tan^2 \theta = 1 \implies \frac{1}{16k^2} - \left(\frac{1}{2h}\right)^2 = 1 \]

### Step 6: Simplify the Equation
Substituting gives:
\[ \frac{1}{16k^2} - \frac{1}{4h^2} = 1 \]
Multiplying through by \(16k^2\) yields:
\[ 1 - 4k^2 = 16k^2h^2 \]
Rearranging gives:
\[ 1 = 4k^2 + 16k^2h^2 \]
Or:
\[ 4k^2 + 16h^2k^2 - 1 = 0 \]

### Step 7: Final Locus Equation
This can be rearranged to:
\[ x^2 - 4y^2 - 16x^2y^2 = 0 \]
Thus, the locus of the midpoint \(M(h, k)\) is:
\[ x^2 - 4y^2 - 16x^2y^2 = 0 \]