Question

Two lines are given by \(x^2-4xy+4y^2+kx-2ky = 0\), then the value of k so that the distance between them is 3 is

• A. \(\sqrt{5}\)
• B. \(3\sqrt{3}\)
• C. \(\sqrt{3}\)
• D. \(3\sqrt{5}\)

Hint:

Factorise the equation as a perfect square plus a linear term.

Answer 1

The Correct Option is D

Step 1: Set $u = x - 2y$:
The equation reads $u^2 + ku = 0$, so $u = 0$ or $u = -k$.

Step 2: Distance:
The lines $x - 2y = 0$ and $x - 2y = -k$ are $\frac{|k|}{\sqrt5}$ apart.
Putting this equal to $3$ gives $|k| = 3\sqrt5$.

Final Answer:
$k = 3\sqrt{5}$, option (D). \[ \boxed{3\sqrt{5}} \]