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(C) The equation of any curve passing through the points of intersection of the given curves is given by $S_1 + \lambda S_2 = 0$,where $S_1 = 2x^2 + 3

Vedclass · Accepted Answer

$90^\circ$

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(C) The equation of any curve passing through the points of intersection of the given curves is given by $S_1 + \lambda S_2 = 0$,where $S_1 = 2x^2 + 3y^2 + 10x$ and $S_2 = 3x^2 + 5y^2 + 16x$.Thus,$(2x^2 + 3y^2 + 10x) + \lambda(3x^2 + 5y^2 + 16x) = 0$.$(2 + 3\lambda)x^2 + (3 + 5\lambda)y^2 + (10 + 16\lambda)x = 0$ ... $(i)$For this equation to represent a pair of straight lines passing through the origin,it must be a homogeneous equation of the second degree. Therefore,the coefficient of $x$ must be zero.$10 + 16\lambda = 0 \Rightarrow \lambda = -\frac{10}{16} = -\frac{5}{8}$.Substituting $\lambda = -\frac{5}{8}$ into equation $(i)$:$(2 + 3(-\frac{5}{8}))x^2 + (3 + 5(-\frac{5}{8}))y^2 = 0$$(\frac{16 - 15}{8})x^2 + (\frac{24 - 25}{8})y^2 = 0$$\frac{1}{8}x^2 - \frac{1}{8}y^2 = 0 \Rightarrow x^2 - y^2 = 0$.This represents the pair of lines $x - y = 0$ and $x + y = 0$.The slopes are $m_1 = 1$ and $m_2 = -1$. Since $m_1 	imes m_2 = -1$,the lines are perpendicular,and the angle between them is $90^\circ$.

The angle between the lines joining the origin to the points of intersection of the curves $2x^2 + 3y^2 + 10x = 0$ and $3x^2 + 5y^2 + 16x = 0$ is

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