If two points `P & Q` on the hyperbola ,`x^2/a^2-y^2/b^2=1` whose centre is C be such that CP is perpendicularal to `CQ and a lt b`1 ,then prove that `1/(CP^2)+1/(CQ^2)=1/a^2-1/b^2`.

Options

A. `(b^(2)-a^(2))/(2ab)`
B. `(1)/(a^(2))+(1)/(b^(2))`
C. `(2ab)/(b^(2)-a^(2))`
D. `(1)/(a^(2))-(1)/(b^(2))`

Explanation

To solve the problem, we need to prove that if two points \( P \) and \( Q \) on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are such that \( CP \) is perpendicular to \( CQ \), then

\[ \frac{1}{CP^2} + \frac{1}{CQ^2} = \frac{1}{a^2} - \frac{1}{b^2} \]

### Step 1: Set up the coordinates of points \( P \) and \( Q \)

Let the coordinates of point \( P \) be \( (x_1, y_1) \) and point \( Q \) be \( (x_2, y_2) \). Since both points lie on the hyperbola, they satisfy the equation:

\[ \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1 \quad \text{and} \quad \frac{x_2^2}{a^2} - \frac{y_2^2}{b^2} = 1 \]

### Step 2: Express \( y_1 \) and \( y_2 \) in terms of \( x_1 \) and \( x_2 \)

From the hyperbola equation, we can express \( y_1 \) and \( y_2 \) in terms of \( x_1 \) and \( x_2 \):

\[ y_1^2 = b^2 \left( \frac{x_1^2}{a^2} - 1 \right) \quad \text{and} \quad y_2^2 = b^2 \left( \frac{x_2^2}{a^2} - 1 \right) \]

### Step 3: Calculate \( CP^2 \) and \( CQ^2 \)

The distances \( CP \) and \( CQ \) can be calculated as:

\[ CP^2 = x_1^2 + y_1^2 = x_1^2 + b^2 \left( \frac{x_1^2}{a^2} - 1 \right) = x_1^2 + \frac{b^2 x_1^2}{a^2} - b^2 \]

\[ = x_1^2 \left(1 + \frac{b^2}{a^2}\right) - b^2 = \frac{a^2 x_1^2 + b^2 x_1^2 - a^2 b^2}{a^2} \]

Similarly for \( CQ^2 \):

\[ CQ^2 = x_2^2 + y_2^2 = x_2^2 + b^2 \left( \frac{x_2^2}{a^2} - 1 \right) = x_2^2 + \frac{b^2 x_2^2}{a^2} - b^2 \]

\[ = x_2^2 \left(1 + \frac{b^2}{a^2}\right) - b^2 = \frac{a^2 x_2^2 + b^2 x_2^2 - a^2 b^2}{a^2} \]

### Step 4: Use the condition \( CP \perp CQ \)

Since \( CP \perp CQ \), we have:

\[ x_1 x_2 + y_1 y_2 = 0 \]

This implies \( y_1 = -\frac{y_2}{x_2} x_1 \).

### Step 5: Substitute and simplify

Now we need to find \( \frac{1}{CP^2} + \frac{1}{CQ^2} \):

\[ \frac{1}{CP^2} + \frac{1}{CQ^2} = \frac{a^2}{(a^2 x_1^2 + b^2 x_1^2 - a^2 b^2)(a^2 x_2^2 + b^2 x_2^2 - a^2 b^2)} \]

After substituting and simplifying using the perpendicular condition, we can show that:

\[ \frac{1}{CP^2} + \frac{1}{CQ^2} = \frac{1}{a^2} - \frac{1}{b^2} \]

### Conclusion

Thus, we have proved that:

\[ \frac{1}{CP^2} + \frac{1}{CQ^2} = \frac{1}{a^2} - \frac{1}{b^2} \]