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

(A) The equation of the curve is $x^{2} + 2xy + 3y^{2} + 4x + 8y - 11 = 0 \dots (1)$ And the line is $y = 3x + 2 \Rightarrow \frac{y - 3x}{2} = 1 \dot

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of the curve is $x^{2} + 2xy + 3y^{2} + 4x + 8y - 11 = 0 \dots (1)$ And the line is $y = 3x + 2 \Rightarrow \frac{y - 3x}{2} = 1 \dots (2)$ Making the equation of the curve homogeneous using equation $(2)$: $x^{2} + 2xy + 3y^{2} + (4x + 8y) \left( \frac{y - 3x}{2} ight) - 11 \left( \frac{y - 3x}{2} ight)^{2} = 0$ $x^{2} + 2xy + 3y^{2} + 2x(y - 3x) + 4y(y - 3x) - \frac{11}{4}(y^{2} - 6xy + 9x^{2}) = 0$ $x^{2} + 2xy + 3y^{2} + 2xy - 6x^{2} + 4y^{2} - 12xy - \frac{11}{4}y^{2} + \frac{66}{4}xy - \frac{99}{4}x^{2} = 0$ Multiplying by $4$: $4x^{2} + 8xy + 12y^{2} + 8xy - 24x^{2} + 16y^{2} - 48xy - 11y^{2} + 66xy - 99x^{2} = 0$ $-119x^{2} + 34xy + 17y^{2} = 0 \Rightarrow 7x^{2} - 2xy - y^{2} = 0 \dots (3)$ Comparing $(3)$ with $ax^{2} + 2hxy + by^{2} = 0$,we get $a = 7, h = -1, b = -1$. The angle $	heta$ between the lines is given by: $	an 	heta = \left| \frac{2\sqrt{h^{2} - ab}}{a + b} ight| = \left| \frac{2\sqrt{(-1)^{2} - (7)(-1)}}{7 - 1} ight| = \frac{2\sqrt{1 + 7}}{6} = \frac{2\sqrt{8}}{6} = \frac{4\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 line $y = 3x + 2$ and the curve $x^{2} + 2xy + 3y^{2} + 4x + 8y - 11 = 0$ is:

Similar Questions

If $\theta$ is the angle between the lines joining the origin to the points of intersection of the curve $2x^2 + 3y^2 = 6$ and the line $x + y = 1$,then $\sin \theta =$

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

The lines joining the points of intersection of the line $x + y = 1$ and the curve $x^2 + y^2 - 2y + \lambda = 0$ to the origin are perpendicular. Then the value of $\lambda$ is:

The pair of straight lines joining the origin to the points of intersection of the line $y = 2\sqrt{2}x + c$ and the circle $x^2 + y^2 = 2$ are at right angles,if

$A$ pair of perpendicular lines passes through the origin and also through the points of intersection of the curve $x^2+y^2=4$ with $x+y=a$,where $a>0$. Then $a$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module