If the normal at `P(asectheta,btantheta)` to the hyperbola `x^2/a^2-y^2/b^2=1` meets the transverse axis in G then minimum length of PG is

Options

A. `(b^(2))/(a)`
B. `|(a)/(b)(a+b)|`
C. `|(a)/(b)(a-b)|`
D. `|(a)/(b)(a-b)|`

Explanation

To solve the problem, we need to find the minimum length of PG, where P is a point on the hyperbola and G is the point where the normal at P intersects the transverse axis. Let's break down the solution step by step.

### Step 1: Identify the Point P
The point P on the hyperbola is given as \( P(a \sec \theta, b \tan \theta) \).

### Step 2: Write the Equation of the Normal
The equation of the normal to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at the point \( P(a \sec \theta, b \tan \theta) \) can be expressed in parametric form as:
\[ a x \cos \theta + b y \sin \theta = a^2 + b^2 \]

### Step 3: Find the Intersection with the Transverse Axis
The transverse axis corresponds to \( y = 0 \). To find the x-coordinate of the point G where the normal intersects the transverse axis, substitute \( y = 0 \) into the normal equation:
\[ a x \cos \theta + b(0) \sin \theta = a^2 + b^2 \]
This simplifies to:
\[ a x \cos \theta = a^2 + b^2 \]
Solving for \( x \), we get:
\[ x = \frac{a^2 + b^2}{a \cos \theta} \]
Thus, the coordinates of point G are:
\[ G\left(\frac{a^2 + b^2}{a \cos \theta}, 0\right) \]

### Step 4: Calculate the Length PG
The length PG can be calculated using the distance formula:
\[ PG = \sqrt{\left(x_G - x_P\right)^2 + \left(y_G - y_P\right)^2} \]
Substituting the coordinates of P and G:
\[ PG = \sqrt{\left(\frac{a^2 + b^2}{a \cos \theta} - a \sec \theta\right)^2 + \left(0 - b \tan \theta\right)^2} \]
This simplifies to:
\[ PG = \sqrt{\left(\frac{a^2 + b^2 - a^2}{a \cos \theta}\right)^2 + \left(-b \tan \theta\right)^2} \]
\[ = \sqrt{\left(\frac{b^2}{a \cos \theta}\right)^2 + \left(-b \tan \theta\right)^2} \]

### Step 5: Simplify the Expression
Now, we simplify the expression:
\[ PG = \sqrt{\frac{b^4}{a^2 \cos^2 \theta} + b^2 \tan^2 \theta} \]
Using the identity \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \):
\[ = \sqrt{\frac{b^4}{a^2 \cos^2 \theta} + \frac{b^2 \sin^2 \theta}{\cos^2 \theta}} \]
\[ = \sqrt{\frac{b^2(b^2 + a^2 \sin^2 \theta)}{a^2 \cos^2 \theta}} \]
\[ = \frac{b}{a} \sqrt{b^2 + a^2 \sin^2 \theta} \]

### Step 6: Find the Minimum Length
To find the minimum length of PG, we need to minimize \( \sqrt{b^2 + a^2 \sin^2 \theta} \). The minimum occurs when \( \sin^2 \theta = 0 \), giving:
\[ PG_{min} = \frac{b}{a} \sqrt{b^2} = \frac{b^2}{a} \]

### Final Answer
Thus, the minimum length of PG is:
\[ \boxed{\frac{b^2}{a}} \]