If the line `4x-5y=0` coincide with one of the lines given by `ax^(2)+2hxy+by^(2)=0`, then

Options

A. `25a-40h-9b=0`
B. `25a+40h-16b=0`
C. `25a-40h+16b=0`
D. `25a+40h+16b=0`

Answer

To solve the problem, we need to determine the conditions under which the line \(4x - 5y = 0\) coincides with one of the lines represented by the equation \(ax^2 + 2hxy + by^2 = 0\).

### Step 1: Rewrite the given line equation
The equation of the line can be rewritten in slope-intercept form:
\[ y = \frac{4}{5}x \]
This indicates that the slope of the line is \(\frac{4}{5}\).

### Step 2: Identify the form of the conic equation
The equation \(ax^2 + 2hxy + by^2 = 0\) represents a pair of straight lines if it can be factored into two linear factors. The general form of a pair of straight lines can be expressed as:
\[ (y - m_1x)(y - m_2x) = 0 \]
where \(m_1\) and \(m_2\) are the slopes of the lines.

### Step 3: Relate the slopes to the coefficients
From the general form, we can derive the relationship between the coefficients \(a\), \(b\), and \(h\) and the slopes \(m_1\) and \(m_2\):
\[ m_1 + m_2 = -\frac{2h}{a} \]
\[ m_1 m_2 = \frac{b}{a} \]

### Step 4: Set one of the slopes equal to the slope of the given line
Since the line \(4x - 5y = 0\) has a slope of \(\frac{4}{5}\), we can set one of the slopes \(m_1\) equal to \(\frac{4}{5}\):
\[ m_1 = \frac{4}{5} \]
Let \(m_2\) be the other slope.

### Step 5: Substitute into the slope equations
Substituting \(m_1\) into the sum of slopes equation:
\[ \frac{4}{5} + m_2 = -\frac{2h}{a} \]
From this, we can express \(m_2\):
\[ m_2 = -\frac{2h}{a} - \frac{4}{5} \]

### Step 6: Substitute into the product of slopes equation
Now substituting \(m_1\) into the product of slopes equation:
\[ \frac{4}{5} m_2 = \frac{b}{a} \]
Substituting \(m_2\) from the previous step:
\[ \frac{4}{5} \left(-\frac{2h}{a} - \frac{4}{5}\right) = \frac{b}{a} \]

### Step 7: Simplify the equation
Multiply both sides by \(5a\) to eliminate the fractions:
\[ 4(-2h - \frac{4a}{5}) = 5b \]
Distributing the \(4\):
\[ -8h - \frac{16a}{5} = 5b \]
Rearranging gives:
\[ 5b + 8h + \frac{16a}{5} = 0 \]

### Conclusion
The condition for the line \(4x - 5y = 0\) to coincide with one of the lines represented by \(ax^2 + 2hxy + by^2 = 0\) is given by:
\[ 5b + 8h + \frac{16a}{5} = 0 \]