Question

If lines represented by the equation $px^2 - qy^2 = 0$ are distinct, then

• A. $p \lt 0$
• B. $p + q = 0$
• C. $pq \gt 0$
• D. $pq = 0$

Hint:

We can rewrite the equation as $px^2 = qy^2 \implies \frac{y^2}{x^2} = \frac{p}{q} \implies \frac{y}{x} = \pm\sqrt{\frac{p}{q}}$. For these lines to be real and distinct, the value under the square root must be strictly positive, meaning $\frac{p}{q} \gt 0$, which is algebraically equivalent to $pq \gt 0$!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Find the condition for px² – qy² = 0 to represent a pair of real and distinct straight lines through the origin.

Step 2: Key Formula or Approach:
For ax² + 2hxy + by² = 0, lines are real and distinct if h² – ab>0. Here a = p, h = 0, b = –q.

Step 3: Detailed Explanation:
h² – ab = 0² – (p)(–q) = pq>0. Thus pq must be strictly positive.

Step 4: Final Answer:
The condition is pq>0, matching option (C).