An ellipse has its centre `C(1,3)` focus at S( 6, 3) and passing through the point `P(4, 7)` then
The product of the lengths of perpendicular segments from the focii on tangent at point P is

Options

A. 20
B. 45
C. 40
D. Cannot be determined

Answer

To solve the problem step by step, we will follow the outlined process in the video transcript and derive the necessary values to find the product of the lengths of perpendicular segments from the foci on the tangent at point P.

### Step 1: Determine the Coordinates of the Foci
Given:
- Center \( C(1, 3) \)
- One focus \( S(6, 3) \)

Using the midpoint formula, we can find the coordinates of the other focus \( S' \):
\[ C = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Let \( S' = (x_1, y_1) \) and \( S = (6, 3) \):
\[ 1 = \frac{x_1 + 6}{2} \quad \Rightarrow \quad x_1 + 6 = 2 \quad \Rightarrow \quad x_1 = -4 \]
\[ 3 = \frac{y_1 + 3}{2} \quad \Rightarrow \quad y_1 + 3 = 6 \quad \Rightarrow \quad y_1 = 3 \]
Thus, the coordinates of the other focus are \( S'(-4, 3) \).

### Step 2: Calculate Distances from Point P to the Foci
Given point \( P(4, 7) \):
1. Distance \( PS' \):
\[ PS' = \sqrt{(4 - (-4))^2 + (7 - 3)^2} = \sqrt{(4 + 4)^2 + (7 - 3)^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \]

2. Distance \( PS \):
\[ PS = \sqrt{(4 - 6)^2 + (7 - 3)^2} = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

### Step 3: Use the Property of Ellipses
From the property of ellipses:
\[ PS + PS' = 2a \]
Substituting the distances we calculated:
\[ 4\sqrt{5} + 2\sqrt{5} = 2a \]
\[ 6\sqrt{5} = 2a \quad \Rightarrow \quad a = 3\sqrt{5} \]

### Step 4: Find the Eccentricity
The distance between the foci \( S \) and \( S' \) is:
\[ d = 2ae \]
Calculating the distance:
\[ d = \sqrt{(6 - (-4))^2 + (3 - 3)^2} = \sqrt{(6 + 4)^2} = \sqrt{10^2} = 10 \]
Setting this equal to \( 2ae \):
\[ 10 = 2(3\sqrt{5})e \quad \Rightarrow \quad 10 = 6\sqrt{5}e \quad \Rightarrow \quad e = \frac{10}{6\sqrt{5}} = \frac{5}{3\sqrt{5}} = \frac{\sqrt{5}}{3} \]

### Step 5: Calculate \( b^2 \)
Using the relationship:
\[ b^2 = a^2(1 - e^2) \]
Calculating \( a^2 \) and \( e^2 \):
\[ a^2 = (3\sqrt{5})^2 = 45 \]
\[ e^2 = \left(\frac{\sqrt{5}}{3}\right)^2 = \frac{5}{9} \]
Thus:
\[ b^2 = 45\left(1 - \frac{5}{9}\right) = 45\left(\frac{4}{9}\right) = 20 \]

### Step 6: Write the Equation of the Ellipse
The equation of the ellipse is:
\[ \frac{(x - 1)^2}{45} + \frac{(y - 3)^2}{20} = 1 \]

### Step 7: Find the Equation of the Tangent at Point P
Using the point-slope form, the slope of the tangent line at point \( P(4, 7) \) is calculated using the derivative of the ellipse equation. The slope is found to be \( -\frac{1}{3} \). The equation of the tangent line is:
\[ x + 3y = 25 \]

### Step 8: Calculate the Lengths of Perpendiculars from Foci to the Tangent
1. For focus \( S(6, 3) \):
\[ L_1 = \frac{|1(6) + 3(3) - 25|}{\sqrt{1^2 + 3^2}} = \frac{|6 + 9 - 25|}{\sqrt{10}} = \frac{10}{\sqrt{10}} = \sqrt{10} \]

2. For focus \( S'(-4, 3) \):
\[ L_2 = \frac{|1(-4) + 3(3) - 25|}{\sqrt{1^2 + 3^2}} = \frac{|-4 + 9 - 25|}{\sqrt{10}} = \frac{20}{\sqrt{10}} = 2\sqrt{10} \]

### Step 9: Calculate the Product of the Lengths
The product of the lengths \( L_1 \) and \( L_2 \):
\[ L_1 \times L_2 = \sqrt{10} \times 2\sqrt{10} = 2 \times 10 = 20 \]

### Final Answer
The product of the lengths of the perpendicular segments from the foci on the tangent at point P is \( \boxed{20} \).