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. $\frac{9}{5}$

Hint:

Logic Tip: The distance between parallel lines given by the general 2nd degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ can be found directly using the formula $d = 2\sqrt{\frac{g^2-ac}{a(a+b)$. Here $a=1, b=4, c=-4, g=3/2$. So $d = 2\sqrt{\frac{9/4 - 1(-4)}{1(1+4) = 2\sqrt{\frac{25/4}{5 = 2\sqrt{5/4} = \sqrt{5}$.

Answer 1

The Correct Option is A

Concept:
The general equation of a second degree is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. For it to represent a pair of parallel straight lines, the second-degree homogeneous part must be a perfect square, meaning $h^2 = ab$. The distance between parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is $d = \frac{|c_1 - c_2|}{\sqrt{a^2+b^2$.
Step 1: Determine the coefficients p and q.
For the equation $x^{2}+4xy+py^{2}+3x+qy-4=0$ to represent parallel lines, the quadratic part $(x^2 + 4xy + py^2)$ must be a perfect square. Comparing with $ax^2 + 2hxy + by^2$, we have $a=1, 2h=4 \implies h=2$, and $b=p$. The condition $h^2 = ab$ gives: $$2^2 = 1 \cdot p \implies p = 4$$ So the quadratic part is $x^2 + 4xy + 4y^2 = (x+2y)^2$. Thus, the parallel lines are of the form $x + 2y + c_1 = 0$ and $x + 2y + c_2 = 0$. Multiplying these gives: $$(x+2y+c_1)(x+2y+c_2) = (x+2y)^2 + (c_1+c_2)(x+2y) + c_1c_2 = 0$$ Comparing this with the given equation $x^2+4xy+4y^2+3x+qy-4=0$: We can see the linear term $3x$ requires $c_1+c_2 = 3$. If the $x$ coefficient is 3, the $y$ coefficient $q$ must be $2(3) = 6$ to maintain the ratio $(x+2y)$. So $q=6$. The constant term is $c_1c_2 = -4$.
Step 2: Find the values of $c_1$ and $c_2$.
We have a system of equations: $$c_1 + c_2 = 3$$ $$c_1c_2 = -4$$ By inspection (or solving the quadratic $t^2 - 3t - 4 = 0$), the roots are $4$ and $-1$. So, $c_1 = 4$ and $c_2 = -1$. The two parallel lines are $x + 2y + 4 = 0$ and $x + 2y - 1 = 0$.
Step 3: Calculate the distance between the lines.
The perpendicular distance $\lambda$ between lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is: $$\lambda = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2$$ Substitute $A=1, B=2, C_1=4, C_2=-1$: $$\lambda = \frac{|4 - (-1)|}{\sqrt{1^2 + 2^2 = \frac{5}{\sqrt{5 = \sqrt{5}$$ The question asks for $\lambda^2$: $$\lambda^2 = (\sqrt{5})^2 = 5$$