Product of perpendiculars from the foci of `x^(2)/4-y^(2)/9=1" to "y=mx+sqrt(4m^2- 9)` where `m gt 3/2` is

Options

A. 4
B. `(36)/(13)`
C. 3
D. 9

Answer

To solve the problem of finding the product of the perpendiculars from the foci of the hyperbola \( \frac{x^2}{4} - \frac{y^2}{9} = 1 \) to the line \( y = mx + \sqrt{4m^2 - 9} \) where \( m > \frac{3}{2} \), we can follow these steps:

### Step 1: Identify the parameters of the hyperbola
The equation of the hyperbola is given as:
\[ \frac{x^2}{4} - \frac{y^2}{9} = 1 \]
From this, we can identify:
- \( a^2 = 4 \) → \( a = 2 \)
- \( b^2 = 9 \) → \( b = 3 \)

### Step 2: Calculate the eccentricity of the hyperbola
The eccentricity \( e \) of the hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \]

### Step 3: Find the foci of the hyperbola
The foci of the hyperbola are located at:
\[ (\pm ae, 0) = \left( \pm 2 \cdot \frac{\sqrt{13}}{2}, 0 \right) = (\pm \sqrt{13}, 0) \]
Thus, the foci are \( (\sqrt{13}, 0) \) and \( (-\sqrt{13}, 0) \).

### Step 4: Set up the line equation
The line equation is given as:
\[ y = mx + \sqrt{4m^2 - 9} \]
We can rewrite it in the standard form \( Ax + By + C = 0 \):
\[ mx - y + \sqrt{4m^2 - 9} = 0 \]
Here, \( A = m \), \( B = -1 \), and \( C = \sqrt{4m^2 - 9} \).

### Step 5: Calculate the perpendicular distances from the foci to the line
The formula for the perpendicular distance \( P \) from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is:
\[ P = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

#### For the focus \( (\sqrt{13}, 0) \):
\[ P_1 = \frac{|m\sqrt{13} - 0 + \sqrt{4m^2 - 9}|}{\sqrt{m^2 + 1}} \]

#### For the focus \( (-\sqrt{13}, 0) \):
\[ P_2 = \frac{|-m\sqrt{13} - 0 + \sqrt{4m^2 - 9}|}{\sqrt{m^2 + 1}} \]

### Step 6: Calculate the product of the perpendicular distances
Now we need to find the product \( P_1 \times P_2 \):
\[ P_1 \times P_2 = \left( \frac{|m\sqrt{13} + \sqrt{4m^2 - 9}|}{\sqrt{m^2 + 1}} \right) \left( \frac{|-m\sqrt{13} + \sqrt{4m^2 - 9}|}{\sqrt{m^2 + 1}} \right) \]

This can be simplified using the identity \( (a + b)(a - b) = a^2 - b^2 \):
\[ P_1 \times P_2 = \frac{(4m^2 - 9) - (13m^2)}{m^2 + 1} \]
\[ = \frac{4m^2 - 13m^2 - 9}{m^2 + 1} = \frac{-9m^2 + 9}{m^2 + 1} = \frac{9(1 - m^2)}{m^2 + 1} \]

### Step 7: Final result
Since \( m > \frac{3}{2} \), we know that \( 1 - m^2 < 0 \), thus:
\[ P_1 \times P_2 = 9 \]
The product of the perpendiculars from the foci of the hyperbola to the line is \( 9 \).