The condition that one of the straight lines given by the equation `ax^(2)+2hxy+by^(2)=0` may coincide with one of those given by the equation `a'x^(2)+2h'xy+b'y^(2)=0` is

Options

A. `(ab'-a'b)^(2)=4(ha'-h'a)(bh'-b'h)`
B. `(ab'=a'b)^(2)=(ha'-h'a)(bh'-b'h)`
C. `(ha'-h'a)=4(ab'-a'b)(bh'-b'h)`
D. `(bh'-b'h)^(2)=4(ab'-a'b)(ha'-h'a)`

Explanation

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

### Step-by-Step Solution:

1. **Assume the Common Line**:
Let the common line be represented by the equation \( y = mx \), where \( m \) is the slope of the line.

2. **Substitute \( y = mx \)**:
Substitute \( y = mx \) into both equations.

For the first equation:
\[ a x^2 + 2h(mx)x + b(mx)^2 = 0 \implies ax^2 + 2hmx^2 + bmx^2 = 0 \]
This simplifies to:
\[ (a + 2hm + bm^2)x^2 = 0 \]

For the second equation:
\[ a'x^2 + 2h'(mx)x + b'(mx)^2 = 0 \implies a'x^2 + 2h'mx^2 + b'm^2x^2 = 0 \]
This simplifies to:
\[ (a' + 2h'm + b'm^2)x^2 = 0 \]

3. **Set Up the Equations**:
From the above simplifications, we have two equations:
\[ a + 2hm + bm^2 = 0 \quad \text{(1)} \]
\[ a' + 2h'm + b'm^2 = 0 \quad \text{(2)} \]

4. **Use Cross Multiplication**:
To find the condition for \( m \), we can use the method of cross multiplication. Rearranging both equations gives:
\[ m^2b + 2hm + a = 0 \quad \text{and} \quad m^2b' + 2h'm + a' = 0 \]
We can express these as quadratic equations in \( m \).

5. **Formulate the Condition**:
For the two equations to have a common solution \( m \), the determinant of the system must be zero. This gives us the condition:
\[ (a' b - a b') = 0 \]
and
\[ (h' a - h a') = 0 \]
and
\[ (b' h - b h') = 0 \]

6. **Final Condition**:
The condition that one of the straight lines given by the first equation coincides with one of those given by the second equation can be summarized as:
\[ (a b' - a' b)^2 = 4(b h' - b' h)(h a' - h' a) \]