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: Use Slopes:
The lines $y=mx$ satisfy $1-2pm-m^2=0$, i.e. $m^2+2pm-1=0$. So $m_1+m_2=-2p$ and $m_1m_2=-1$.

Step 2: Bisector Angle:
If the lines make angles $\theta_1,\theta_2$ with the x-axis, a bisector makes angle $\varphi=(\theta_1+\theta_2)/2$. So $\tan2\varphi=\tan(\theta_1+\theta_2)=\dfrac{m_1+m_2}{1-m_1m_2}=\dfrac{-2p}{2}=-p$.

Step 3: Use the Given Bisector Pair:
A bisector slope $n$ satisfies $n^2+2qn-1=0$, so $1-n^2=2qn$. Then $\tan2\varphi=\dfrac{2n}{1-n^2}=\dfrac{2n}{2qn}=\dfrac1q$. Setting $\dfrac1q=-p$ gives $pq=-1$. Option (D).

Final Answer:
Option (D). \[ \boxed{\text{(D) } pq+1=0} \]