The line 3x + 4y - 5 = 0 cuts the curve 2x^2 | vedclass

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(C) The given equation of the line is $3x + 4y - 5 = 0$,which can be written as $\frac{3x + 4y}{5} = 1$ $(i)$.The equation of the curve is $2x^2 + 3y^

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$\frac{\pi}{2}$

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(C) The given equation of the line is $3x + 4y - 5 = 0$,which can be written as $\frac{3x + 4y}{5} = 1$ $(i)$.The equation of the curve is $2x^2 + 3y^2 = 5$ (ii).To find the angle $\angle AOB$,we homogenize the equation of the curve using the line equation:$2x^2 + 3y^2 = 5(1)^2$$2x^2 + 3y^2 = 5\left(\frac{3x + 4y}{5}ight)^2$$2x^2 + 3y^2 = 5\left(\frac{9x^2 + 16y^2 + 24xy}{25}ight)$$2x^2 + 3y^2 = \frac{9x^2 + 16y^2 + 24xy}{5}$$10x^2 + 15y^2 = 9x^2 + 16y^2 + 24xy$$x^2 - 24xy - y^2 = 0$This is a homogeneous equation of the second degree representing the pair of lines $OA$ and $OB$. The general form is $ax^2 + 2hxy + by^2 = 0$. Here,$a = 1$,$b = -1$,and $h = -12$.The angle $	heta$ between the lines is given by $	an 	heta = \left|\frac{2\sqrt{h^2 - ab}}{a + b}ight|$.Since $a + b = 1 + (-1) = 0$,the lines are perpendicular.Therefore,$\angle AOB = 90^{\circ} = \frac{\pi}{2}$.

The line $3x + 4y - 5 = 0$ cuts the curve $2x^2 + 3y^2 = 5$ at $A$ and $B$. If $O$ is the origin,then $\angle AOB =$

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