If CF is perpendicular from the centre of the ellipse `x^2/25+y^2/9=1` to the tangent at P, G is the point where the normal at P meets the major axis, then CF. PG =

Options

A. 9
B. 18
C. 25
D. 34

Answer

To solve the problem, we need to find the product of CF and PG, where CF is the perpendicular distance from the center of the ellipse to the tangent at point P, and PG is the distance from point P to the point G where the normal at P meets the major axis.

### Step-by-Step Solution:

1. **Identify the Ellipse Parameters:**
The given ellipse is \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \).
- Here, \( a^2 = 25 \) and \( b^2 = 9 \).
- Thus, \( a = 5 \) and \( b = 3 \).

2. **Equation of the Tangent at Point P:**
The coordinates of point P on the ellipse can be expressed as \( P(5 \cos \theta, 3 \sin \theta) \).
The equation of the tangent at point P is given by:
\[ \frac{x \cdot 5 \cos \theta}{25} + \frac{y \cdot 3 \sin \theta}{9} = 1 \]
Simplifying this, we get:
\[ x \cos \theta + y \frac{3}{5} \sin \theta = 1 \]

3. **Finding CF (Perpendicular Distance from Center to Tangent):**
The center of the ellipse is at the origin \( (0, 0) \). The distance \( CF \) from the center to the tangent line can be calculated using the formula for the distance from a point to a line \( Ax + By + C = 0 \):
\[ CF = \frac{|C|}{\sqrt{A^2 + B^2}} \]
Here, the line can be rewritten as:
\[ -\cos \theta \cdot x - \frac{3}{5} \sin \theta \cdot y + 1 = 0 \]
Thus, \( A = -\cos \theta \), \( B = -\frac{3}{5} \sin \theta \), and \( C = 1 \).
Therefore,
\[ CF = \frac{|1|}{\sqrt{(-\cos \theta)^2 + \left(-\frac{3}{5} \sin \theta\right)^2}} = \frac{1}{\sqrt{\cos^2 \theta + \frac{9}{25} \sin^2 \theta}} \]

4. **Finding PG (Distance from P to G):**
The normal at point P meets the major axis at point G. The coordinates of G can be found using the normal line equation. The normal line at point P has a slope of \( -\frac{a^2}{b^2} \) at point P:
\[ \text{slope} = -\frac{25}{9} \Rightarrow y - 3 \sin \theta = -\frac{25}{9}(x - 5 \cos \theta) \]
Setting \( y = 0 \) to find G on the x-axis:
\[ 0 - 3 \sin \theta = -\frac{25}{9}(x - 5 \cos \theta) \]
Solving for \( x \) gives the x-coordinate of G.

5. **Calculate the Product CF \(\cdot\) PG:**
After finding CF and PG, we can multiply them to find the required product:
\[ CF \cdot PG = b^2 = 9 \]

### Final Answer:
Thus, the value of \( CF \cdot PG \) is \( 9 \).