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 angle ACB=90◦, then `1/(CA)^(2)+1/((CB)^(2))=`

Options

A. `7/12`
B. `12/7`
C. `25/144`
D. `144/25`

Answer

To solve the problem, we need to find the value of \( \frac{1}{(CA)^2} + \frac{1}{(CB)^2} \) for points \( A \) and \( B \) on the ellipse defined by the equation \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \) where the angle \( ACB = 90^\circ \).

### Step-by-step Solution:

1. **Identify the center and parameters of the ellipse**:
The given ellipse is \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \).
- The center \( C \) of the ellipse is at the origin \( (0, 0) \).
- The semi-major axis \( a = 4 \) (since \( a^2 = 16 \)) and the semi-minor axis \( b = 3 \) (since \( b^2 = 9 \)).

2. **Parametric representation of points on the ellipse**:
The points \( A \) and \( B \) on the ellipse can be represented parametrically as:
- \( A = (4 \cos \alpha, 3 \sin \alpha) \)
- \( B = (4 \cos \beta, 3 \sin \beta) \)

3. **Condition for angle \( ACB = 90^\circ \)**:
For \( \angle ACB = 90^\circ \), the slopes of lines \( CA \) and \( CB \) must satisfy the condition:
\[ m_1 \cdot m_2 = -1 \]
where \( m_1 \) is the slope of line \( CA \) and \( m_2 \) is the slope of line \( CB \).

- The slope \( m_1 = \frac{3 \sin \alpha}{4 \cos \alpha} = \frac{3}{4} \tan \alpha \)
- The slope \( m_2 = \frac{3 \sin \beta}{4 \cos \beta} = \frac{3}{4} \tan \beta \)

Thus, we have:
\[ \left(\frac{3}{4} \tan \alpha\right) \left(\frac{3}{4} \tan \beta\right) = -1 \]
Simplifying gives:
\[ \tan \alpha \tan \beta = -\frac{16}{9} \]

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

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

Using the identity \( \tan \alpha \tan \beta = -\frac{16}{9} \), we can simplify this expression.

6. **Final Calculation**:
From the previous steps, we can derive that:
\[ \frac{1}{(CA)^2} + \frac{1}{(CB)^2} = \frac{1}{a^2} + \frac{1}{b^2} \]
where \( a^2 = 16 \) and \( b^2 = 9 \).

Thus:
\[ \frac{1}{(CA)^2} + \frac{1}{(CB)^2} = \frac{1}{16} + \frac{1}{9} \]

Finding a common denominator (which is 144):
\[ = \frac{9}{144} + \frac{16}{144} = \frac{25}{144} \]

### Final Answer:
\[ \frac{1}{(CA)^2} + \frac{1}{(CB)^2} = \frac{25}{144} \]