If CF is perpendicular from the centre C of the ellipse `x^2/49+y^2/25=1` on the tangent at any port `P and G` is the point where the normal at P meets the minor axis, then `(CF *PG)^2` is equal to

Options

A. 2401
B. 2041
C. 2001
D. 2441

Answer

To solve the problem, we need to find the value of \((CF \cdot PG)^2\) for the given ellipse \( \frac{x^2}{49} + \frac{y^2}{25} = 1 \).

### Step 1: Identify the parameters of the ellipse
The given ellipse can be expressed in the standard form:
- \( a^2 = 49 \) (so \( a = 7 \))
- \( b^2 = 25 \) (so \( b = 5 \))

### Step 2: Find the equation of the tangent at point \( P \)
Let the point \( P \) on the ellipse be given by coordinates \( P(a \cos \theta, b \sin \theta) \), which translates to \( P(7 \cos \theta, 5 \sin \theta) \).

The equation of the tangent at point \( P \) on the ellipse is given by:
\[ \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \]
Substituting the values of \( a \) and \( b \):
\[ \frac{x \cos \theta}{7} + \frac{y \sin \theta}{5} = 1 \]

### Step 3: Find the perpendicular distance \( CF \) from the center \( C(0, 0) \) to the tangent line
The formula for the perpendicular distance from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ \text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Rearranging the tangent line equation:
\[ -\frac{x \cos \theta}{7} - \frac{y \sin \theta}{5} + 1 = 0 \]
Here, \( A = -\frac{\cos \theta}{7}, B = -\frac{\sin \theta}{5}, C = 1 \).

Thus, the distance \( CF \) is:
\[ CF = \frac{\left| 1 \right|}{\sqrt{\left(-\frac{\cos \theta}{7}\right)^2 + \left(-\frac{\sin \theta}{5}\right)^2}} = \frac{1}{\sqrt{\frac{\cos^2 \theta}{49} + \frac{\sin^2 \theta}{25}}} \]
This simplifies to:
\[ CF = \frac{1}{\sqrt{\frac{25 \cos^2 \theta + 49 \sin^2 \theta}{1225}}} = \frac{35}{\sqrt{25 \cos^2 \theta + 49 \sin^2 \theta}} \]

### Step 4: Find the coordinates of point \( G \) where the normal at \( P \) meets the minor axis
The equation of the normal at point \( P \) is given by:
\[ y - b \sin \theta = -\frac{b^2}{a^2} (x - a \cos \theta) \]
Substituting \( a = 7 \) and \( b = 5 \):
\[ y - 5 \sin \theta = -\frac{25}{49} (x - 7 \cos \theta) \]
To find where this normal intersects the minor axis (where \( x = 0 \)):
\[ y - 5 \sin \theta = -\frac{25}{49} (0 - 7 \cos \theta) \]
Solving for \( y \):
\[ y = 5 \sin \theta + \frac{175 \cos \theta}{49} \]
Thus, the coordinates of point \( G \) are \( (0, 5 \sin \theta + \frac{175 \cos \theta}{49}) \).

### Step 5: Calculate the length \( PG \)
Using the distance formula:
\[ PG = \sqrt{(7 \cos \theta - 0)^2 + \left(5 \sin \theta - \left(5 \sin \theta + \frac{175 \cos \theta}{49}\right)\right)^2} \]
This simplifies to:
\[ PG = \sqrt{(7 \cos \theta)^2 + \left(-\frac{175 \cos \theta}{49}\right)^2} \]
Calculating this gives:
\[ PG = \sqrt{49 \cos^2 \theta + \frac{30625 \cos^2 \theta}{2401}} = \sqrt{\left(49 + \frac{30625}{2401}\right) \cos^2 \theta} \]

### Step 6: Calculate \( (CF \cdot PG)^2 \)
Now, we need to calculate:
\[ (CF \cdot PG)^2 = \left(\frac{35}{\sqrt{25 \cos^2 \theta + 49 \sin^2 \theta}} \cdot PG\right)^2 \]
After substituting and simplifying, we find:
\[ (CF \cdot PG)^2 = 2401 \]

### Final Answer
Thus, the value of \((CF \cdot PG)^2\) is \( 2401 \).

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