The tangent at P on the hyperbola `(x^(2))/(a^(2)) -(y^(2))/(b^(2))=1` meets one of the asymptote in Q. Then the locus of the mid-point of PQ is

Options

A. `3((x^(2))/(a^(2))-(y^(2))/(b^(2)))=4`
B. `(x^(2))/(a^(2))-(y^(2))/(b^(2)) =2`
C. `(x^(2))/(a^(2)) -(y^(2))/(b^(2)) =(1)/(2)`
D. `4((x^(2))/(a^(2))-(y^(2))/(b^(2)))=3`

Explanation

To find the locus of the midpoint of PQ, where P is a point on the hyperbola and Q is the point where the tangent at P meets one of the asymptotes, we can follow these steps:

### Step 1: Write the equation of the hyperbola
The equation of the hyperbola is given as:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Find the coordinates of point P
Let the coordinates of point P on the hyperbola be represented in parametric form as:
\[ P = (a \sec \theta, b \tan \theta) \]

### Step 3: Write the equation of the tangent at point P
The equation of the tangent at point P can be derived using the point-slope form:
\[ \frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1 \]
This can be rearranged to:
\[ x \sec \theta - y \frac{a}{b} \tan \theta = a \]

### Step 4: Write the equation of one asymptote
The equations of the asymptotes for the hyperbola are given by:
\[ \frac{x}{a} + \frac{y}{b} = 1 \quad \text{(we will use this asymptote)} \]

### Step 5: Find the coordinates of point Q
To find the coordinates of point Q where the tangent meets the asymptote, we can substitute the equation of the tangent into the equation of the asymptote.
Let the coordinates of Q be:
\[ Q = (x_Q, y_Q) \]
From the asymptote equation, we can express \(y_Q\) in terms of \(x_Q\):
\[ y_Q = b - \frac{b}{a} x_Q \]

### Step 6: Set up the equations to find the midpoint M
Let M be the midpoint of PQ. The coordinates of M can be calculated as:
\[ M = \left(\frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}\right) \]

### Step 7: Substitute the coordinates of P and Q
Substituting the coordinates of P and Q into the midpoint formula:
\[ M = \left(\frac{a \sec \theta + x_Q}{2}, \frac{b \tan \theta + y_Q}{2}\right) \]

### Step 8: Express x_Q and y_Q in terms of \(\theta\)
Using the equations derived previously, we can express \(x_Q\) and \(y_Q\) in terms of \(\theta\):
\[ x_Q = a \sec \theta - b \tan \theta \]
\[ y_Q = b \tan \theta - a \sec \theta \]

### Step 9: Find the locus of M
Now, we can express the coordinates of M in terms of \(\theta\):
\[ h = \frac{a \sec \theta + (a \sec \theta - b \tan \theta)}{2} = a \sec \theta - \frac{b \tan \theta}{2} \]
\[ k = \frac{b \tan \theta + (b \tan \theta - a \sec \theta)}{2} = b \tan \theta - \frac{a \sec \theta}{2} \]

### Step 10: Eliminate \(\theta\)
We can derive two equations from the expressions for \(h\) and \(k\):
1. \( \frac{h}{a} + \frac{k}{b} = \frac{\sec \theta + \tan \theta}{2} \)
2. \( \frac{h}{a} - \frac{k}{b} = \frac{3}{2}(\sec \theta - \tan \theta) \)

Multiplying these two equations gives us the required locus:
\[ \frac{h^2}{a^2} - \frac{k^2}{b^2} = \frac{3}{4} \]

### Final Locus Equation
Multiplying through by 4 gives:
\[ 4\frac{h^2}{a^2} - 4\frac{k^2}{b^2} = 3 \]
Replacing \(h\) with \(x\) and \(k\) with \(y\):
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = \frac{3}{4} \]

Thus, the locus of the midpoint M is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = \frac{3}{4} \]