Question

If the distance between the parallel lines given by the equation $x^2 + 4xy + py^2 + 3x + qy - 4 = 0$ is $\lambda$, then $\lambda^2 =$

• A. 5
• B. $\sqrt{5}$
• C. 25
• D. $\frac{9}{5}$

Answer 1

Answer: (A)

5

Answer 2

Solution:

For the given quadratic to represent two parallel lines, the quadratic part must be a perfect square. Write

$$x^2+4xy+py^2=(x+2y)^2=\;x^2+4xy+4y^2.$$

Thus, we require

$$p=4.$$

The equation becomes

$$(x+2y)^2+3x+qy-4=0.$$

Express $3x+qy$ in terms of $(x+2y)$. Write the equation in factorized form:

$$(x+2y+k_1)(x+2y+k_2)=0.$$

Expanding,

$$(x+2y)^2+(k_1+k_2)(x+2y)+k_1k_2=0.$$

Comparing with the original equation, we have:

$$k_1+k_2=3 \quad \text{and} \quad k_1k_2=-4.$$

The two lines are:

$$x+2y+k_1=0 \quad \text{and} \quad x+2y+k_2=0.$$

The distance between parallel lines $L_1: x+2y+c_1=0$ and $L_2: x+2y+c_2=0$ is given by:

$$\lambda = \frac{|c_1-c_2|}{\sqrt{1^2+2^2}} = \frac{|c_1-c_2|}{\sqrt{5}}.$$

Noting that $|c_1-c_2|=\sqrt{(k_1-k_2)^2}$ and using the identity

$$(k_1-k_2)^2 = (k_1+k_2)^2-4k_1k_2,$$

we substitute $k_1+k_2=3$ and $k_1k_2=-4$:

$$(k_1-k_2)^2 = 3^2 - 4(-4) = 9+16 = 25.$$

Thus,

$$|k_1-k_2| = 5,$$ $$\lambda = \frac{5}{\sqrt{5}} = \sqrt{5}\quad \Rightarrow\quad \lambda^2=5.$$

Answer: Option (A) 5

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Explanation (minimal):

• Require $p=4$ for a perfect square: $(x+2y)^2$.
• Factorize to $(x+2y+k_1)(x+2y+k_2)=0$ with $k_1+k_2=3$ and $k_1k_2=-4$.
• Compute $|k_1-k_2|=\sqrt{25}=5$.
• Distance $\lambda=\frac{5}{\sqrt{5}}=\sqrt{5}$ so $\lambda^2=5$.