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 pair of lines joining the origin to the points of intersection of the line $x-y=2$ with the curve $5x^2+12xy-8y^2+8x-4y+12=0$ are equally inclined to the pair of lines

If $f(x, y) = 0$ is the combined equation of the lines joining the origin to the points where the line $4x - 6y - 2 = 0$ meets the curve $3x^2 - 4xy + 5y^2 - 2x + y - 6 = 0$,then $\frac{f(1, -1)}{f(-1, -1)} = $

The distance between the pair of parallel lines represented by $x^2+2xy+y^2-8ax-8ay-9a^2=0$ is $...$ units.

Let $3x^2+8xy-3y^2=0$ represent the lines $L_1, L_2$ and $3x^2+8xy-3y^2+2x-4y-1=0$ represent the lines $L_3, L_4$. Let $L$ be the line joining the points of intersection of $L_1, L_3$ and $L_2, L_4$. Then,the area (in sq units) of the triangle formed by $L$ with the coordinate axes is

The acute angle between the pair of straight lines joining the origin to the points of intersection of the line $x+y-1=0$ with the pair of straight lines $k x^2+8 x y-3 y^2+2 x-4 y-1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module