The product of the perpendiculars drawn from the loci of the ellipse `x^2/9+y^2/25=1` upon the tangent to it at the point `(3/2, (5sqrt3)/2)` is

Options

A. `3sqrt13`
B. 9
C. `189//13`
D. 18

Answer

To solve the problem, we need to find the product of the perpendiculars drawn from the foci of the ellipse \( \frac{x^2}{9} + \frac{y^2}{25} = 1 \) to the tangent at the point \( \left( \frac{3}{2}, \frac{5\sqrt{3}}{2} \right) \).

### Step-by-Step Solution:

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

2. **Find the foci of the ellipse**:
The foci of the ellipse are located at \( (0, \pm c) \), where \( c = \sqrt{b^2 - a^2} \).
\[ c = \sqrt{25 - 9} = \sqrt{16} = 4 \]
Therefore, the foci are at \( (0, 4) \) and \( (0, -4) \).

3. **Determine the equation of the tangent line**:
The equation of the tangent to the ellipse at the point \( (x_1, y_1) = \left( \frac{3}{2}, \frac{5\sqrt{3}}{2} \right) \) can be found using the formula:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
Substituting \( a^2 = 9 \), \( b^2 = 25 \), \( x_1 = \frac{3}{2} \), and \( y_1 = \frac{5\sqrt{3}}{2} \):
\[ \frac{3}{2} \cdot \frac{x}{9} + \frac{5\sqrt{3}}{2} \cdot \frac{y}{25} = 1 \]
Simplifying this gives the equation of the tangent.

4. **Use the property of the ellipse**:
According to the property of the ellipse, the product of the perpendiculars drawn from the foci to any tangent is given by:
\[ L_1 \cdot L_2 = b^2 \]
Here, \( b^2 = 25 \).

5. **Final answer**:
Therefore, the product of the perpendiculars drawn from the foci to the tangent at the given point is:
\[ L_1 \cdot L_2 = 25 \]

### Summary:
The product of the perpendiculars drawn from the foci of the ellipse to the tangent at the point \( \left( \frac{3}{2}, \frac{5\sqrt{3}}{2} \right) \) is \( 25 \).