If ` 4x-3y+k=0` touches the ellipse ` 5x^(2)+9y^(2)=45` , then k is equal to

Options

A. ` pm 3 sqrt(21)`
B. `3 sqrt(21)`
C. `-3 sqrt(21)`
D. `2 sqrt(21)`

Explanation

To find the value of \( k \) such that the line \( 4x - 3y + k = 0 \) touches the ellipse \( 5x^2 + 9y^2 = 45 \), we can follow these steps:

### Step 1: Convert the equation of the ellipse to standard form
The given equation of the ellipse is:
\[ 5x^2 + 9y^2 = 45 \]
To convert it to standard form, we divide the entire equation by 45:
\[ \frac{5x^2}{45} + \frac{9y^2}{45} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} + \frac{y^2}{5} = 1 \]
Thus, the semi-major axis \( a = 3 \) (since \( \sqrt{9} = 3 \)) and the semi-minor axis \( b = \sqrt{5} \).

**Hint:** To convert an ellipse equation to standard form, divide by the constant on the right side to make it equal to 1.

### Step 2: Rewrite the line equation in slope-intercept form
The line equation is:
\[ 4x - 3y + k = 0 \]
Rearranging this to solve for \( y \):
\[ 3y = 4x + k \implies y = \frac{4}{3}x + \frac{k}{3} \]
Here, the slope \( m = \frac{4}{3} \) and the y-intercept \( c = \frac{k}{3} \).

**Hint:** To find the slope and intercept of a line, rearrange the equation into the form \( y = mx + c \).

### Step 3: Use the condition for tangency
For the line to be tangent to the ellipse, the distance from the center of the ellipse to the line must equal the semi-minor axis. The distance \( d \) from the center \( (0, 0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|C|}{\sqrt{A^2 + B^2}} \]
In our case, \( A = 4 \), \( B = -3 \), and \( C = k \):
\[ d = \frac{|k|}{\sqrt{4^2 + (-3)^2}} = \frac{|k|}{\sqrt{16 + 9}} = \frac{|k|}{5} \]
Since the line touches the ellipse, this distance must equal the semi-minor axis \( b = \sqrt{5} \):
\[ \frac{|k|}{5} = \sqrt{5} \]

**Hint:** The distance from a point to a line can be calculated using the formula involving the coefficients of the line equation.

### Step 4: Solve for \( k \)
From the equation:
\[ |k| = 5\sqrt{5} \]
Thus, \( k \) can be either:
\[ k = 5\sqrt{5} \quad \text{or} \quad k = -5\sqrt{5} \]

**Hint:** The absolute value equation can yield both positive and negative solutions.

### Final Result
The possible values of \( k \) are:
\[ k = \pm 5\sqrt{5} \]