If the equation x^2+2 sqrt{2} xy + 2y^2 + 4x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given equation is $x^2 + 2 \sqrt{2} xy + 2y^2 + 4x + 4 \sqrt{2}y + 1 = 0$. This can be rewritten as $(x + \sqrt{2}y)^2 + 4(x + \sqrt{2}y) + 1

Vedclass · Accepted Answer

$2$ units

Vedclass · Answer

(B) The given equation is $x^2 + 2 \sqrt{2} xy + 2y^2 + 4x + 4 \sqrt{2}y + 1 = 0$. This can be rewritten as $(x + \sqrt{2}y)^2 + 4(x + \sqrt{2}y) + 1 = 0$. Let $t = x + \sqrt{2}y$. Then the equation becomes $t^2 + 4t + 1 = 0$. Solving for $t$ using the quadratic formula,$t = \frac{-4 \pm \sqrt{16 - 4}}{2} = -2 \pm \sqrt{3}$. Thus,the two parallel lines are $x + \sqrt{2}y + 2 - \sqrt{3} = 0$ and $x + \sqrt{2}y + 2 + \sqrt{3} = 0$. The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. Here,$A = 1$,$B = \sqrt{2}$,$C_1 = 2 - \sqrt{3}$,and $C_2 = 2 + \sqrt{3}$. $d = \frac{|(2 - \sqrt{3}) - (2 + \sqrt{3})|}{\sqrt{1^2 + (\sqrt{2})^2}} = \frac{|-2 \sqrt{3}|}{\sqrt{3}} = \frac{2 \sqrt{3}}{\sqrt{3}} = 2$ units. Therefore,the correct option is $B$.

If the equation $x^2+2 \sqrt{2} xy + 2y^2 + 4x + 4 \sqrt{2}y + 1 = 0$ represents a pair of parallel straight lines,find the distance between them.

Similar Questions

The equation of the pair of straight lines joining the origin to the points of intersection of the curve $x^2 + y^2 = 4$ and the line $y - x = 2$ is

Let $L$ be the line joining the origin to the point of intersection of the lines represented by $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$. If $L$ is perpendicular to the line $kx + y + 3 = 0$,then $k$ is equal to

The line $x+2y=k$ meets the curve $2x^2-2xy+3y^2+2x-y-1=0$ at two points $A$ and $B$. Let $O$ be the origin. If the line segments $OA$ and $OB$ are perpendicular to each other,then $k=$

The lines joining the origin to the points of intersection of the curves $ax^2 + 2hxy + by^2 + 2gx = 0$ and $a'x^2 + 2h'xy + b'y^2 + 2g'x = 0$ will be mutually perpendicular,if

The lines joining the origin to the points of intersection of the line $3x - 2y = 1$ and the curve $3x^2 + 5xy - 3y^2 + 2x + 3y = 0$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module