The tangent at 'p' on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` cuts the major axis in T and PN is the perpendicular to the x-axis, C being centre then CN.CT =

Options

A. a
B. `a^(2)`
C. b
D. `b^(2)`

Answer

To solve the problem, we need to find the product of the distances CN and CT, where C is the center of the ellipse, N is the foot of the perpendicular from point P on the ellipse to the x-axis, and T is the point where the tangent at P intersects the major axis of the ellipse.

### Step-by-Step Solution:

1. **Identify the Ellipse and Points**:
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The center of the ellipse C is at the origin (0, 0). Let P be a point on the ellipse represented as \( P(x_1, y_1) \).

2. **Equation of the Tangent**:
The equation of the tangent to the ellipse at point P can be expressed as:
\[ \frac{x \cdot x_1}{a^2} + \frac{y \cdot y_1}{b^2} = 1 \]

3. **Finding Point T**:
To find the point T where the tangent intersects the major axis (x-axis), we set \( y = 0 \) in the tangent equation:
\[ \frac{x \cdot x_1}{a^2} + 0 = 1 \implies x = \frac{a^2}{x_1} \]
Thus, the coordinates of point T are \( T\left(\frac{a^2}{x_1}, 0\right) \).

4. **Finding Distance CN**:
The distance CN is the distance from the center C(0, 0) to the point N, which is the projection of P onto the x-axis. The coordinates of N are \( N(x_1, 0) \). Therefore, the distance CN is:
\[ CN = |x_1 - 0| = |x_1| \]

5. **Finding Distance CT**:
The distance CT is the distance from the center C(0, 0) to the point T. The coordinates of T are \( T\left(\frac{a^2}{x_1}, 0\right) \). Thus, the distance CT is:
\[ CT = \left|\frac{a^2}{x_1} - 0\right| = \frac{a^2}{|x_1|} \]

6. **Calculating the Product CN \cdot CT**:
Now, we calculate the product CN \cdot CT:
\[ CN \cdot CT = |x_1| \cdot \frac{a^2}{|x_1|} = a^2 \]

### Final Result:
Thus, we conclude that:
\[ CN \cdot CT = a^2 \]