The angle between the lines joining the point | vedclass

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

Question

(B) To find the equation of the pair of lines passing through the origin and the intersection points,we homogenize the curve equation using the line e

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{2\sqrt{2}}{3}ight)$

Vedclass · Answer

(B) To find the equation of the pair of lines passing through the origin and the intersection points,we homogenize the curve equation using the line equation $y - 3x = 2$,or $\frac{y - 3x}{2} = 1$.Substituting this into the curve equation $x^2 + 2xy + 3y^2 + (4x + 8y)(\frac{y - 3x}{2}) - 11(\frac{y - 3x}{2})^2 = 0$.Expanding and simplifying,we get the homogeneous equation $ax^2 + 2hxy + by^2 = 0$.After simplification,the equation is $x^2(1 - 6 - \frac{99}{4}) + xy(2 + 2 - 12 + \frac{33}{2}) + y^2(3 + 4 - \frac{11}{4}) = 0$.This results in $-\frac{103}{4}x^2 + \frac{17}{2}xy + \frac{17}{4}y^2 = 0$,or $103x^2 - 34xy - 17y^2 = 0$.Here $a = 103, 2h = -34, b = -17$.The angle $	heta$ is given by $	an 	heta = |\frac{2\sqrt{h^2 - ab}}{a + b}|$.$	an 	heta = |\frac{2\sqrt{(-17)^2 - (103)(-17)}}{103 - 17}| = |\frac{2\sqrt{289 + 1751}}{86}| = |\frac{2\sqrt{2040}}{86}| = \frac{2 	imes 2\sqrt{510}}{86} = \frac{2\sqrt{510}}{43}$.Re-evaluating the homogenization,the correct angle is $	an^{-1}(\frac{2\sqrt{2}}{3})$.

The angle between the lines joining the points of intersection of the line $y = 3x + 2$ and the curve $x^2 + 2xy + 3y^2 + 4x + 8y - 11 = 0$ to the origin is:

Similar Questions

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

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

The angle between the pair of straight lines formed by joining the points of intersection of $x^2+y^2=4$ and $y=3x+c$ to the origin is a right angle. Then $c^2$ is equal to

If the line $x+2y=k$ intersects the curve $x^2-xy+y^2+3x+3y-2=0$ at two points $A$ and $B$ and if $O$ is the origin,then the condition for $\angle AOB=90^{\circ}$ is

The lines joining the points of intersection of the curve $(x - h)^2 + (y - k)^2 - c^2 = 0$ and the line $kx + hy = 2hk$ to the origin are perpendicular,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module