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 $8x^2 + 8xy + 2y^2 + 26x + 13y + 15 = 0$ represents a pair of parallel straight lines. The distance between them is

$A$ pair of lines drawn through the origin forms a right-angled isosceles triangle with the line $2x + 3y = 6$,having the right angle at the origin. The area (in sq. units) of the triangle thus formed is

The combined equation of the bisectors of the angles between the lines joining the origin to the points of intersection of the curve $x^2+y^2+xy+x+3y+1=0$ and the line $x+y+2=0$ is

The acute angle formed between the lines joining the origin to the points of intersection of the curves $x^2 + y^2 - 2x - 1 = 0$ and $x + y = 1$ is

If the pair of lines joining the origin to the points of intersection of the line $x+y=1$ with the curve $x^2+y^2+2hxy+gx+fy+1=0$ are at right angles,then the point $(g, f)$ lies on the line

Vedclass Test Series

Exam Paper Generator

Online Exam Module