If theta is the acute angle between the lines | vedclass

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

Question

(B) The equation of the curve is $x^2+xy+y^2+x+3y+1=0$ and the line is $x+y+2=0$. To homogenize the equation of the curve using the line,we write the

Vedclass · Accepted Answer

$\frac{1}{\sqrt{5}}$

Vedclass · Answer

(B) The equation of the curve is $x^2+xy+y^2+x+3y+1=0$ and the line is $x+y+2=0$. To homogenize the equation of the curve using the line,we write the line as $\frac{x+y}{-2}=1$. Substituting this into the curve equation: $x^2+xy+y^2+(x+3y)(\frac{x+y}{-2}) + 1(\frac{x+y}{-2})^2=0$. Multiplying by $4$ to clear the denominators: $4x^2+4xy+4y^2-2(x^2+4xy+3y^2)+(x^2+2xy+y^2)=0$. $4x^2+4xy+4y^2-2x^2-8xy-6y^2+x^2+2xy+y^2=0$. $3x^2-2xy-y^2=0$. Comparing this with $ax^2+2hxy+by^2=0$,we have $a=3, h=-1, b=-1$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight|$. $	an 	heta = \left| \frac{2\sqrt{(-1)^2-3(-1)}}{3-1} ight| = \left| \frac{2\sqrt{4}}{2} ight| = 2$. Since $	an 	heta = 2$,we have $\cos 	heta = \frac{1}{\sqrt{1+	an^2 	heta}} = \frac{1}{\sqrt{1+4}} = \frac{1}{\sqrt{5}}$.

If $\theta$ is the acute angle between the lines joining the origin to the points of intersection of the curve $x^2+xy+y^2+x+3y+1=0$ and the straight line $x+y+2=0$,then $\cos \theta=$

Similar Questions

What is the distance between the lines represented by the equation $x^2 + 2\sqrt{2}xy + 2y^2 + 4x + 4\sqrt{2}y + 1 = 0$?

If the distance between the parallel lines given by the equation $x^2+4xy+4y^2+3x+6y-4=0$ is $\lambda$,then $\lambda^2=$

Find the value$(s)$ of $k$ such that the distance between the two parallel lines represented by $(x-2y)^2 + k(x-2y) = 0$ is $3$ units.

The equation of the pair of lines joining the origin to the points of intersection of two circles $x^2+y^2-4x+8y+5=0$ and $x^2+y^2+2x+4y-3=0$ is

If the pair of lines joining the origin and the points of intersection of the line $ax+by=1$ and the curve $x^2+y^2-x-y-1=0$ are at right angles,then the locus of the point $(a, b)$ is a circle of radius

Vedclass Test Series

Exam Paper Generator

Online Exam Module