The angle between the lines joining the origi | vedclass

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

Question

(D) The equation of the line is $y - 3x = 2$,which can be written as $\frac{y - 3x}{2} = 1$.To homogenize the curve equation $x^2 + 2xy + 3y^2 + 4x +

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The equation of the line is $y - 3x = 2$,which can be written as $\frac{y - 3x}{2} = 1$.To homogenize the curve equation $x^2 + 2xy + 3y^2 + 4x + 8y - 11 = 0$,we substitute $1$ with $\frac{y - 3x}{2}$:$x^2 + 2xy + 3y^2 + (4x + 8y)\left(\frac{y - 3x}{2}ight) - 11\left(\frac{y - 3x}{2}ight)^2 = 0$.Multiplying by $4$ to clear the denominator:$4x^2 + 8xy + 12y^2 + 2(4x + 8y)(y - 3x) - 11(y - 3x)^2 = 0$.$4x^2 + 8xy + 12y^2 + 2(4xy - 12x^2 + 8y^2 - 24xy) - 11(y^2 - 6xy + 9x^2) = 0$.$4x^2 + 8xy + 12y^2 - 24x^2 - 40xy + 16y^2 - 99x^2 + 66xy - 11y^2 = 0$.$-119x^2 + 34xy + 17y^2 = 0$,which simplifies to $17y^2 + 34xy - 119x^2 = 0$,or $y^2 + 2xy - 7x^2 = 0$.For a homogeneous equation $ax^2 + 2hxy + by^2 = 0$,the angle $	heta$ is given by $	an 	heta = \left|\frac{2\sqrt{h^2 - ab}}{a + b}ight|$.Here $a = -7, h = 1, b = 1$.$	an 	heta = \left|\frac{2\sqrt{1^2 - (-7)(1)}}{-7 + 1}ight| = \left|\frac{2\sqrt{8}}{-6}ight| = \frac{2(2\sqrt{2})}{6} = \frac{2\sqrt{2}}{3}$.Therefore,$	heta = 	an^{-1}\left(\frac{2\sqrt{2}}{3}ight)$.

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

Similar Questions

If the distance between the lines represented by $(x - 2y)^2 + k(x - 2y) = 0$ is $3$ units and $k > 0$, then the equations of the lines are:

$A$ pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve $x^2+y^2=4$ with $x+y=a$. The set containing the value of $a$ is

Let the line $2x - 3y - 1 = 0$ intersect the curve $x^2 + 2xy + 5y^2 + 2x + 3y - 1 = 0$ at distinct points $A$ and $B$. If $O$ is the origin,then $\cos \angle AOB =$

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

Consider the lines $L_1 \equiv 4x + 5y - 6 = 0$,$L_2 \equiv 2x + 3y - 4 = 0$,and $L_3 \equiv 3x - y + 2 = 0$. If the line $L_1 = 0$ intersects the lines $L_2 = 0$ and $L_3 = 0$ at the points $A$ and $B$ respectively,then the combined equation of lines $OA$ and $OB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module