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(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

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$	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:

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