Question

The equation of second degree$x^2+2\sqrt{2}xy+2y^2+4x+4\sqrt{2}y+1=0$ represents a pair of straight lines. The distance between them is

• A. $2\sqrt[]{3}$
• B. $2\sqrt[]{5}$
• C. $2$
• D. $10$

Answer 1

The correct option is C

$2$

Explanation for correct option:

Finding the distance:

We know that the distance between pair of straight lines in the equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ is

$\text{Distance}=2\sqrt{\frac{(g^2–ac)}{a(a+b)}}\text{or}2\sqrt{\frac{(f^2–bc)}{b(a+b)}}$

The given equation is $x^2+2\sqrt{2}xy+2y^2+4x+4\sqrt{2}y+1=0$

Comparing this with the general equation ${\mathbf{ax}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{hxy}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{by}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{gx}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{fy}\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$, we get

$$\begin{gathered} a=1, \\ h=\sqrt{2}, \\ b=2, \\ g=2, \\ c=1 \end{gathered}$$

Now, finding the distance using the formula,

$\begin{array}{rcl}\text{Distance}& =& 2\sqrt[]{\frac{(g^2–ac)}{a(a+b)}}\\ & =& 2\sqrt[]{\frac{(2^2–1)}{1(1+2)}}\\ & =& 2\sqrt[]{\frac{3}{3}}\\ \text{Distance}& =& 2\end{array}$

Hence, option (C) is the correct answer.