C is the centre of the ellipse `x^(2)/16+y^(2)/9=1` and A and B are two points on the ellipse such that
`/_ACB=90^@`, then `1/(CA)^(2)+1/((CB)^(2))=`

Options

A. (a) `7/12`
B. (b) `12/7`
C. (c) `25/144`
D. (d) `144/25`

Answer

To solve the problem, we need to find the value of \( \frac{1}{CA^2} + \frac{1}{CB^2} \) given that \( C \) is the center of the ellipse defined by the equation \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \) and that \( \angle ACB = 90^\circ \).

### Step-by-Step Solution:

1. **Identify the Center of the Ellipse**:
The center \( C \) of the ellipse is at the origin, \( (0, 0) \).

2. **Parametric Coordinates of Points A and B**:
The parametric equations for points \( A \) and \( B \) on the ellipse are:
\[ A = (4 \cos \alpha, 3 \sin \alpha) \]
\[ B = (4 \cos \beta, 3 \sin \beta) \]
Here, \( a = 4 \) and \( b = 3 \) are the semi-major and semi-minor axes of the ellipse, respectively.

3. **Calculate the Slopes of Lines AC and BC**:
The slope \( m_1 \) of line \( AC \) is given by:
\[ m_1 = \frac{3 \sin \alpha - 0}{4 \cos \alpha - 0} = \frac{3 \sin \alpha}{4 \cos \alpha} \]
The slope \( m_2 \) of line \( BC \) is:
\[ m_2 = \frac{3 \sin \beta - 0}{4 \cos \beta - 0} = \frac{3 \sin \beta}{4 \cos \beta} \]

4. **Use the Perpendicularity Condition**:
Since \( AC \) is perpendicular to \( BC \), we have:
\[ m_1 \cdot m_2 = -1 \]
Substituting the slopes:
\[ \left(\frac{3 \sin \alpha}{4 \cos \alpha}\right) \cdot \left(\frac{3 \sin \beta}{4 \cos \beta}\right) = -1 \]
Simplifying gives:
\[ \frac{9 \sin \alpha \sin \beta}{16 \cos \alpha \cos \beta} = -1 \]
Rearranging leads to:
\[ \sin \alpha \sin \beta = -\frac{16}{9} \cos \alpha \cos \beta \]

5. **Finding CA and CB**:
Using the distance formula, we find:
\[ CA^2 = (4 \cos \alpha)^2 + (3 \sin \alpha)^2 = 16 \cos^2 \alpha + 9 \sin^2 \alpha \]
\[ CB^2 = (4 \cos \beta)^2 + (3 \sin \beta)^2 = 16 \cos^2 \beta + 9 \sin^2 \beta \]

6. **Calculate \( \frac{1}{CA^2} + \frac{1}{CB^2} \)**:
We need to find:
\[ \frac{1}{CA^2} + \frac{1}{CB^2} = \frac{1}{16 \cos^2 \alpha + 9 \sin^2 \alpha} + \frac{1}{16 \cos^2 \beta + 9 \sin^2 \beta} \]

7. **Using the Identity for \( \tan \alpha \tan \beta \)**:
From the previous steps, we can express \( CA^2 \) and \( CB^2 \) in terms of \( \tan \alpha \) and \( \tan \beta \) using the relationship derived from the slopes.

8. **Final Calculation**:
After substituting and simplifying, we find:
\[ \frac{1}{CA^2} + \frac{1}{CB^2} = \frac{1}{9} + \frac{1}{16} = \frac{16 + 9}{144} = \frac{25}{144} \]

Thus, the final answer is:
\[ \frac{1}{CA^2} + \frac{1}{CB^2} = \frac{25}{144} \]