Key Concepts and Formulas

• Condition for a Pair of Straight Lines: The general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if and only if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
• Quadratic Equation with Repeated Roots: The quadratic equation $Ax^2 + Bx + C = 0$ has repeated roots if and only if its discriminant is zero, i.e., $B^2 - 4AC = 0$.
• Intersection on the y-axis: A point on the y-axis has an x-coordinate of 0.

Step-by-Step Solution

Step 1: Find the equation representing the intersection of the pair of lines with the y-axis.

• Explanation: Since the pair of lines intersects on the y-axis, we substitute $x = 0$ into the given equation to find the y-coordinates of the intersection points.
• Working: $a(0)^2 + 2h(0)y + by^2 + 2g(0) + 2fy + c = 0$ $by^2 + 2fy + c = 0 \quad \text{(Equation 1)}$
• Interpretation: Equation 1 is a quadratic equation in $y$. Its roots represent the y-coordinates where the pair of lines intersects the y-axis.

Step 2: Apply the condition for a unique intersection point on the y-axis.

• Explanation: The problem states that the pair of lines intersects on the y-axis, meaning at a single, unique point. Therefore, Equation 1 must have repeated roots. We apply the discriminant condition for repeated roots.
• Working: For the quadratic equation $by^2 + 2fy + c = 0$ to have repeated roots, its discriminant must be zero: $(2f)^2 - 4(b)(c) = 0$ $4f^2 - 4bc = 0$ Dividing by 4: $f^2 - bc = 0$ $f^2 = bc \quad \text{(Equation 2)}$

Step 3: Use the general condition for a pair of straight lines.

• Explanation: We use the condition that the given equation represents a pair of straight lines, which is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. We substitute the condition derived in Step 2 ($f^2 = bc$) into this general condition.
• Working: The general condition for a pair of straight lines is: $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ Substitute $f^2 = bc$ into the equation: $abc + 2fgh - a(bc) - bg^2 - ch^2 = 0$ $abc + 2fgh - abc - bg^2 - ch^2 = 0$ The terms $abc$ and $-abc$ cancel out: $2fgh - bg^2 - ch^2 = 0$ Rearranging the terms: $2fgh = bg^2 + ch^2$

Step 4: Conclusion.

• Explanation: We compare the derived condition with the given options.
• Working: The derived condition $2fgh = bg^2 + ch^2$ matches option (A).

Common Mistakes & Tips

• Memorize the General Condition: The condition $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ is crucial.
• Understand the difference between "intersects the y-axis" and "intersects on the y-axis": The latter implies a unique intersection point, leading to repeated roots.
• Careful with Algebra: Ensure correct substitutions and simplification to avoid errors.

Summary

By substituting $x=0$ into the equation of the pair of lines, we obtain a quadratic equation in $y$. Since the pair of lines intersects on the y-axis, this quadratic must have repeated roots, giving $f^2 = bc$. Combining this with the general condition for a pair of straight lines, $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$, we arrive at the condition $2fgh = bg^2 + ch^2$.

The final answer is $\boxed{2fgh = bg^2 + ch^2}$, which corresponds to option (A).