Find the condition that one of the lines given by `ax^2+2hxy+by^2=0` may coincide with one of the lines given by `a' x^2 +2h'xy+b'y^2=0`

Answer

To find the condition under which one of the lines given by the equation \( ax^2 + 2hxy + by^2 = 0 \) coincides with one of the lines given by the equation \( a'x^2 + 2h'xy + b'y^2 = 0 \), we will follow these steps:

### Step 1: Assume the form of the line
Assume that the line that coincides can be expressed in the form \( y = mx \).

### Step 2: Substitute into the first equation
Substituting \( y = mx \) into the first equation \( ax^2 + 2hxy + by^2 = 0 \):
\[ a x^2 + 2h x(mx) + b(mx)^2 = 0 \]
This simplifies to:
\[ a x^2 + 2hm x^2 + bm^2 x^2 = 0 \]
Factoring out \( x^2 \) (assuming \( x \neq 0 \)):
\[ (a + 2hm + bm^2) = 0 \]
Let this be Equation (1):
\[ bm^2 + 2hm + a = 0 \]

### Step 3: Substitute into the second equation
Now, substitute \( y = mx \) into the second equation \( a'x^2 + 2h'xy + b'y^2 = 0 \):
\[ a' x^2 + 2h' x(mx) + b'(mx)^2 = 0 \]
This simplifies to:
\[ a' x^2 + 2h'm x^2 + b'm^2 x^2 = 0 \]
Factoring out \( x^2 \):
\[ (a' + 2h'm + b'm^2) = 0 \]
Let this be Equation (2):
\[ b'm^2 + 2h'm + a' = 0 \]

### Step 4: Solve the two equations
Now we have two quadratic equations in \( m \):
1. \( bm^2 + 2hm + a = 0 \) (Equation 1)
2. \( b'm^2 + 2h'm + a' = 0 \) (Equation 2)

For the lines to coincide, the two equations must have at least one common solution for \( m \). This means their discriminants must be equal, or they must be proportional.

### Step 5: Set up the condition
The condition for the two equations to have a common solution can be derived from the relationship between their coefficients. The condition can be expressed as:
\[ \frac{b}{b'} = \frac{2h}{2h'} = \frac{a}{a'} \]

### Final Condition
Thus, the condition that one of the lines given by \( ax^2 + 2hxy + by^2 = 0 \) may coincide with one of the lines given by \( a'x^2 + 2h'xy + b'y^2 = 0 \) is:
\[ a b' - a' b = 0, \quad 2h b' - 2h' b = 0 \]