Question

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

• A. $pq > 0$
• B. $pq < 0$
• C. $pq = 0$
• D. $p + q = 0$

Answer 1

Answer: (A)

A

Answer 2

The given equation is $px^2 - qy^2 = 0$. This equation represents a pair of straight lines passing through the origin.

We can rewrite the equation as $px^2 = qy^2$.

Case 1: If $p=0$ or $q=0$.

• If $p=0$, the equation becomes $-qy^2 = 0$. If $q \neq 0$, this simplifies to $y^2 = 0$, which means $y=0$. This is the equation of the x-axis, a single line.
• If $q=0$, the equation becomes $px^2 = 0$. If $p \neq 0$, this simplifies to $x^2 = 0$, which means $x=0$. This is the equation of the y-axis, a single line.

In both these scenarios ($p=0$ or $q=0$), the equation represents a single line, not two distinct lines. Therefore, for distinct lines, we must have $p \neq 0$ and $q \neq 0$.

Case 2: If $p \neq 0$ and $q \neq 0$.

We can write the equation as $y^2 = \frac{p}{q} x^2$.

Taking the square root on both sides, we get: $y = \pm \sqrt{\frac{p}{q}} x$

For these two lines to be real and distinct, the term under the square root, $\frac{p}{q}$, must be a positive real number.

• If $\frac{p}{q} < 0$, then $\sqrt{\frac{p}{q}}$ would be an imaginary number, meaning the lines would be imaginary (they would only intersect at the origin in the real plane).
• If $\frac{p}{q} = 0$, this would imply $p=0$ (since $q \neq 0$), which we have already excluded as it leads to a single line.

Therefore, for two distinct real lines, we must have $\frac{p}{q} > 0$.

The condition $\frac{p}{q} > 0$ implies that $p$ and $q$ must have the same sign:

• If $p > 0$ and $q > 0$, then $pq > 0$.
• If $p < 0$ and $q < 0$, then $pq > 0$.

Thus, the condition for the lines to be distinct (real and distinct) is $pq > 0$.

Alternative Method (Using General Equation of Pair of Straight Lines):

The general equation of a pair of straight lines passing through the origin is $Ax^2 + 2Hxy + By^2 = 0$.

The given equation is $px^2 - qy^2 = 0$.

Comparing the coefficients:

$A = p$ $H = 0$ $B = -q$

For the lines to be real and distinct, the condition is $H^2 - AB > 0$.

Substituting the values:

$(0)^2 - (p)(-q) > 0$ $0 + pq > 0$ $pq > 0$

Both methods yield the same result.