Let the line 2x - 3y - 1 = 0 intersect the cu | vedclass

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

Question

(D) The equation of the line is $2x - 3y = 1$. We homogenize the equation of the curve $x^2 + 2xy + 5y^2 + 2x + 3y - 1 = 0$ using the line equation: $

Vedclass · Accepted Answer

$\frac{3 \sqrt{2}}{7}$

Vedclass · Answer

(D) The equation of the line is $2x - 3y = 1$. We homogenize the equation of the curve $x^2 + 2xy + 5y^2 + 2x + 3y - 1 = 0$ using the line equation: $x^2 + 2xy + 5y^2 + (2x + 3y)(1) - (1)^2 = 0$ Substituting $1 = 2x - 3y$: $x^2 + 2xy + 5y^2 + (2x + 3y)(2x - 3y) - (2x - 3y)^2 = 0$ $x^2 + 2xy + 5y^2 + (4x^2 - 9y^2) - (4x^2 - 12xy + 9y^2) = 0$ $x^2 + 2xy + 5y^2 + 4x^2 - 9y^2 - 4x^2 + 12xy - 9y^2 = 0$ $x^2 + 14xy - 13y^2 = 0$ This represents the pair of lines $OA$ and $OB$. Comparing with $ax^2 + 2hxy + by^2 = 0$,we get $a = 1$,$2h = 14 \Rightarrow h = 7$,and $b = -13$. The angle $	heta$ between the lines is given by $	an 	heta = \frac{2\sqrt{h^2 - ab}}{|a + b|}$. $	an 	heta = \frac{2\sqrt{7^2 - (1)(-13)}}{|1 - 13|} = \frac{2\sqrt{49 + 13}}{12} = \frac{2\sqrt{62}}{12} = \frac{\sqrt{62}}{6}$. Using $\cos 	heta = \frac{1}{\sqrt{1 + 	an^2 	heta}} = \frac{1}{\sqrt{1 + \frac{62}{36}}} = \frac{1}{\sqrt{\frac{98}{36}}} = \frac{6}{\sqrt{98}} = \frac{6}{7\sqrt{2}} = \frac{3\sqrt{2}}{7}$.

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 $\cos \angle AOB =$

Similar Questions

If the lines joining the origin to the points of intersection of the line $x+y=k$ and the curve $x^2+y^2-2x-4y+2=0$ are at right angles,then the sum of all the possible values of $k$ is

The distance between the two lines represented by the pair of straight lines $9x^2 - 24xy + 16y^2 + 3x - 4y - 6 = 0$ is:

The equation of the bisectors of the angles 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 $x+y+2=0$ is

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 $\cos \theta=$

The distance between the parallel lines represented by the equation $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module