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:
We are given a homogeneous second-degree equation $px^2 - qy^2 = 0$ that represents a pair of straight lines passing through the origin. We need to find the specific condition under which these lines are real and distinct.

Step 2: Key Formula or Approach:
For any standard homogeneous pair of straight lines $ax^2 + 2hxy + by^2 = 0$, the condition for the lines to be real and distinct is: $$h^2 - ab \gt 0$$ Mapping our parameters from the given equation: $a = p$, $h = 0$, and $b = -q$.

Step 3: Detailed Explanation:
Substitute our mapped coefficients into the distinct condition formula: $$h^2 - ab \gt 0$$ $$(0)^2 - (p)(-q) \gt 0$$ $$0 - (-pq) \gt 0$$ $$pq \gt 0$$ Thus, for the lines to be distinct, the product of the coefficients $p$ and $q$ must be strictly positive.

Step 4: Final Answer:
The required condition is $pq \gt 0$, matching option (C).

Step 4: Final Answer:
The matching condition is $pq \gt 0$, which corresponds to option (C).