Question

If the combined equation of angle bisectors of the lines \(x^2-2pxy-y^2 = 0\) is \(x^2-2qxy-y^2 = 0\), then which of the following is true?

• A. \(2p+q = 0\)
• B. \(2p+3q = 0\)
• C. \(pq = 1\)
• D. \(pq+1 = 0\)

Hint:

Use the bisector equation (x^2 - y^2)/(a - b) = xy/h for the pair ax^2 + 2hxy + by^2 = 0.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For the pair \(ax^2+2hxy+by^2=0\), the combined equation of the angle bisectors is
\[ \frac{x^2-y^2}{a-b}=\frac{xy}{h} \]

Step 2: Identify the Coefficients:
For \(x^2-2pxy-y^2=0\): \(a=1,\ b=-1,\ 2h=-2p\), so \(h=-p\).

Step 3: Form the Bisector Equation:
\[ \frac{x^2-y^2}{2}=\frac{xy}{-p} \Rightarrow x^2-y^2=-\frac{2}{p}xy \Rightarrow x^2+\frac{2}{p}xy-y^2=0 \]

Step 4: Compare With the Given Form:
The bisector pair is given as \(x^2-2qxy-y^2=0\). Matching the xy coefficients: \(-2q=\dfrac{2}{p}\), so \(pq=-1\), or \(pq+1=0\).
This matches option (D). Options (A) and (B) are linear relations in p and q, while the actual relation is a product, and option (C) has the wrong sign.

Final Answer:
The relation is \(pq+1=0\), option (D). \[ \boxed{\text{(D) } pq+1=0} \]