If one of the two lines `6x^(2)+xy-y^(2)=0` coincides with one of the two lines `3x^(2)-axy+y^(2)=0` ten

Answer

To solve the problem, we need to find the value of \( a \) such that one line from the first equation coincides with one line from the second equation. The equations given are:

1. \( 6x^2 + xy - y^2 = 0 \)
2. \( 3x^2 - axy + y^2 = 0 \)

### Step 1: Factor the first equation

The first equation can be factored. We rewrite it as:

\[ 6x^2 + xy - y^2 = 0 \]

We can rearrange it to group terms:

\[ 6x^2 + 3xy - 2xy - y^2 = 0 \]

Now, we can factor by grouping:

\[ 3x(2x + y) - y(2x + y) = 0 \]

This gives us:

\[ (3x - y)(2x + y) = 0 \]

Thus, the two lines represented by the first equation are:

1. \( y = 3x \) (slope = 3)
2. \( y = -2x \) (slope = -2)

### Step 2: Analyze the second equation

Now, we consider the second equation:

\[ 3x^2 - axy + y^2 = 0 \]

We can also factor this equation. We rearrange it as:

\[ 3x^2 - axy + y^2 = 0 \]

Dividing through by \( x^2 \) (assuming \( x \neq 0 \)) gives:

\[ 3 - a\frac{y}{x} + \left(\frac{y}{x}\right)^2 = 0 \]

Let \( m = \frac{y}{x} \), then we have:

\[ m^2 - am + 3 = 0 \]

### Step 3: Condition for coinciding lines

For the lines to coincide, the quadratic equation must have a double root. This occurs when the discriminant is zero:

\[ D = b^2 - 4ac = 0 \]

Here, \( a = 1 \), \( b = -a \), and \( c = 3 \). Thus, we have:

\[ (-a)^2 - 4(1)(3) = 0 \]

This simplifies to:

\[ a^2 - 12 = 0 \]

### Step 4: Solve for \( a \)

Solving for \( a \):

\[ a^2 = 12 \]

Taking the square root of both sides gives:

\[ a = \sqrt{12} \quad \text{or} \quad a = -\sqrt{12} \]

Thus, we have:

\[ a = 2\sqrt{3} \quad \text{or} \quad a = -2\sqrt{3} \]

### Final Answer

The values of \( a \) such that one line from the first equation coincides with one line from the second equation are:

\[ a = 2\sqrt{3} \quad \text{or} \quad a = -2\sqrt{3} \]