The tangent to `x^(2)//a^(2)+y^(2)//b^(2)=1` meets the major and minor axes in P and Q respectively, then `a^(2)//CP^(2)+b^(2)//CQ^(2)=`

Answer

To solve the problem, we need to find the value of \( \frac{a^2}{CP^2} + \frac{b^2}{CQ^2} \) where \( CP \) and \( CQ \) are the distances from the center of the ellipse to the points where the tangent meets the major and minor axes.

### Step-by-Step Solution:

1. **Equation of the Ellipse**:
The equation of the ellipse is given as:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

2. **Tangent to the Ellipse**:
The equation of the tangent to the ellipse at an angle \( \theta \) can be expressed as:
\[ \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \]

3. **Finding Points P and Q**:
- **Point P (on the major axis)**: Set \( y = 0 \) in the tangent equation to find \( x \):
\[ \frac{x \cos \theta}{a} = 1 \implies x = \frac{a}{\cos \theta} \]
Thus, the coordinates of point P are \( \left(\frac{a}{\cos \theta}, 0\right) \).
- **Point Q (on the minor axis)**: Set \( x = 0 \) in the tangent equation to find \( y \):
\[ \frac{y \sin \theta}{b} = 1 \implies y = \frac{b}{\sin \theta} \]
Thus, the coordinates of point Q are \( \left(0, \frac{b}{\sin \theta}\right) \).

4. **Calculating Distances CP and CQ**:
- The distance \( CP \) from the center to point P is:
\[ CP = \frac{a}{\cos \theta} \]
- The distance \( CQ \) from the center to point Q is:
\[ CQ = \frac{b}{\sin \theta} \]

5. **Calculating \( CP^2 \) and \( CQ^2 \)**:
- \( CP^2 = \left(\frac{a}{\cos \theta}\right)^2 = \frac{a^2}{\cos^2 \theta} \)
- \( CQ^2 = \left(\frac{b}{\sin \theta}\right)^2 = \frac{b^2}{\sin^2 \theta} \)

6. **Finding \( \frac{a^2}{CP^2} + \frac{b^2}{CQ^2} \)**:
- Substitute the values of \( CP^2 \) and \( CQ^2 \):
\[ \frac{a^2}{CP^2} = \frac{a^2}{\frac{a^2}{\cos^2 \theta}} = \cos^2 \theta \]
\[ \frac{b^2}{CQ^2} = \frac{b^2}{\frac{b^2}{\sin^2 \theta}} = \sin^2 \theta \]
- Therefore, we have:
\[ \frac{a^2}{CP^2} + \frac{b^2}{CQ^2} = \cos^2 \theta + \sin^2 \theta \]

7. **Using the Pythagorean Identity**:
- We know that:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]

8. **Final Result**:
Thus, the final answer is:
\[ \frac{a^2}{CP^2} + \frac{b^2}{CQ^2} = 1 \]