Key Concepts and Formulas

• Equation of Angle Bisectors: For a pair of straight lines $ax^2 + 2hxy + by^2 = 0$, the equation of the angle bisectors is given by $\frac{x^2 - y^2}{a-b} = \frac{xy}{h}$.
• Identical Pair of Lines: Two homogeneous second-degree equations $a_1x^2 + 2h_1xy + b_1y^2 = 0$ and $a_2x^2 + 2h_2xy + b_2y^2 = 0$ represent the same pair of lines if and only if their coefficients are proportional: $\frac{a_1}{a_2} = \frac{h_1}{h_2} = \frac{b_1}{b_2}$.

Step-by-Step Solution

Step 1: Identify the coefficients for $L_2$

We are given $L_2: x^2 - 2qxy - y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$, we identify the coefficients as:

• $a = 1$
• $2h = -2q \implies h = -q$
• $b = -1$

Step 2: Find the equation of the angle bisectors of $L_2$

We use the formula for the angle bisectors: $\frac{x^2 - y^2}{a-b} = \frac{xy}{h}$. Substituting the coefficients $a=1$, $b=-1$, and $h=-q$ into the formula gives: $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-q}$ Simplifying the denominator: $\frac{x^2 - y^2}{2} = \frac{xy}{-q}$ Cross-multiplying to eliminate the denominators: $-q(x^2 - y^2) = 2xy$ Expanding the left side: $-qx^2 + qy^2 = 2xy$ Rearranging the terms to get a standard form: $qx^2 + 2xy - qy^2 = 0 \quad (*)$ This equation represents the angle bisectors of the pair of lines given by $L_2$.

Step 3: Apply the Identity Condition

The problem states that the angle bisectors of $L_2$ are the same as $L_1$. Therefore, equation (*) must represent the same pair of lines as $L_1: x^2 - 2pxy - y^2 = 0$. For two equations to represent the same lines, their coefficients must be proportional. Thus, comparing $qx^2 + 2xy - qy^2 = 0$ and $x^2 - 2pxy - y^2 = 0$, we have: $\frac{q}{1} = \frac{2}{-2p} = \frac{-q}{-1}$

Step 4: Solve for the Relationship between $p$ and $q$

We can equate any two parts of the proportionality. Let's use the first two parts: $\frac{q}{1} = \frac{2}{-2p}$ $q = -\frac{1}{p}$ Multiplying both sides by $p$ gives: $pq = -1$ We can also verify this by equating the first and third parts: $\frac{q}{1} = \frac{-q}{-1}$ $q = q$ This doesn't directly give the relationship between $p$ and $q$, but it confirms consistency. Let's equate the second and third parts: $\frac{2}{-2p} = \frac{-q}{-1}$ $-\frac{1}{p} = q$ $pq = -1$ Thus, we get the same relationship.

Common Mistakes & Tips

• Sign Errors: Pay close attention to signs when substituting values into the angle bisector formula and when comparing coefficients. A simple sign error can lead to an incorrect answer.
• Coefficient of $xy$: Remember that the general form is $ax^2 + 2hxy + by^2 = 0$, so the coefficient of $xy$ is $2h$, not $h$.
• Understanding Proportionality: If two homogeneous second-degree equations represent the same pair of lines, it means their coefficients are proportional, not necessarily equal.

Summary

By finding the equation of the angle bisectors of the second pair of lines ($L_2$) and using the condition that this pair is identical to the first pair of lines ($L_1$), we derived the relationship between $p$ and $q$. This relationship is given by $pq = -1$.

Final Answer

The final answer is $\boxed{pq = -1}$, which corresponds to option (A).