Let the curve x^2+2y^2=2 intersect the line x | vedclass

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(B) The equation of the curve is $x^2+2y^2=2$ and the line is $x+y=1$. To find the lines $OP$ and $OQ$,we homogenize the equation of the curve using t

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$4$

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(B) The equation of the curve is $x^2+2y^2=2$ and the line is $x+y=1$. To find the lines $OP$ and $OQ$,we homogenize the equation of the curve using the line equation: $x^2+2y^2=2(1)^2$ $x^2+2y^2=2(x+y)^2$ $x^2+2y^2=2(x^2+y^2+2xy)$ $x^2+2y^2=2x^2+2y^2+4xy$ $x^2+4xy=0$ This represents a pair of lines passing through the origin. Comparing this with the general form $ax^2+2hxy+by^2=0$,we have $a=1$,$h=2$,and $b=0$. The angle $	heta$ between these lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight|$. Substituting the values: $	an 	heta = \left| \frac{2\sqrt{2^2-(1)(0)}}{1+0} ight| = \left| \frac{2\sqrt{4}}{1} ight| = 4$.

Let the curve $x^2+2y^2=2$ intersect the line $x+y=1$ at two points $P$ and $Q$,and let $O$ be the origin. If $\theta$ is the acute angle between the lines $OP$ and $OQ$,then $\tan \theta=$

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