The locus of the middle points of the portions of the tangents of the hyperbola. `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` included between the axes is

Options

A. `a^(2)x^(2)-b^(2)y^(2)=x^(2)y^(2)`
B. `b^(2)x^(2)-a^(2)y^(2)=x^(2)y^(2)`
C. `b^(2)x^(2)-a^(2)y^(2)=4x^(2)y^(2)`
D. `a^(2)x^(2)-b^(2)y^(2)=4x^(2)y^(2)`

Answer

To find the locus of the midpoints of the portions of the tangents of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) included between the axes, we can follow these steps:

### Step 1: Write the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola at a point \((x_1, y_1)\) is given by:
\[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]
For our purposes, we can consider the point on the hyperbola as \((a \sec \theta, b \tan \theta)\).

### Step 2: Substitute the point into the tangent equation
Substituting \(x_1 = a \sec \theta\) and \(y_1 = b \tan \theta\) into the tangent equation gives:
\[ \frac{x(a \sec \theta)}{a^2} - \frac{y(b \tan \theta)}{b^2} = 1 \]
This simplifies to:
\[ \frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1 \]

### Step 3: Find the points where the tangent intersects the axes
- **Intersection with the x-axis**: Set \(y = 0\):
\[ \frac{x \sec \theta}{a} = 1 \implies x = a \cos \theta \]
Thus, the point of intersection with the x-axis is \((a \cos \theta, 0)\).

- **Intersection with the y-axis**: Set \(x = 0\):
\[ -\frac{y \tan \theta}{b} = 1 \implies y = -b \cot \theta \]
Thus, the point of intersection with the y-axis is \((0, -b \cot \theta)\).

### Step 4: Find the midpoint of the segment between the intersections
The midpoint \(M(h, k)\) of the segment connecting the points \((a \cos \theta, 0)\) and \((0, -b \cot \theta)\) is given by:
\[ h = \frac{a \cos \theta + 0}{2} = \frac{a \cos \theta}{2} \]
\[ k = \frac{0 + (-b \cot \theta)}{2} = -\frac{b \cot \theta}{2} \]

### Step 5: Express \(\sec \theta\) and \(\tan \theta\) in terms of \(h\) and \(k\)
From the expressions for \(h\) and \(k\):
\[ \cos \theta = \frac{2h}{a} \quad \text{and} \quad \cot \theta = -\frac{2k}{b} \]
Using the identity \(\sec^2 \theta - \tan^2 \theta = 1\):
\[ \sec^2 \theta = \frac{1}{\cos^2 \theta} = \frac{a^2}{(2h)^2} = \frac{a^2}{4h^2} \]
\[ \tan^2 \theta = \frac{1}{\cot^2 \theta} = \frac{b^2}{(2k)^2} = \frac{b^2}{4k^2} \]
Substituting these into the identity gives:
\[ \frac{a^2}{4h^2} - \frac{b^2}{4k^2} = 1 \]
Multiplying through by \(4h^2k^2\) yields:
\[ a^2k^2 - b^2h^2 = 4h^2k^2 \]

### Step 6: Rearranging to find the locus
Rearranging gives:
\[ a^2k^2 - 4h^2k^2 - b^2h^2 = 0 \]
This is the equation of the locus of the midpoints of the tangents of the hyperbola.

### Final Result
The locus of the midpoints of the portions of the tangents of the hyperbola included between the axes is:
\[ \frac{a^2}{4h^2} - \frac{b^2}{4k^2} = 1 \]