For the variable t, the locus of the points o | vedclass

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(D) Given equations of lines are:$x-2y=t$  $(i)$$x+2y=\frac{1}{t}$  $(ii)$Multiplying equations $(i)$ and $(ii)$, we get:$(x-2y)(x+2y) = t 	imes \fra

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the hyperbola with centre at the origin and one focus $\left(\frac{\sqrt{5}}{2}, 0\right)$

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(D) Given equations of lines are:$x-2y=t$  $(i)$$x+2y=\frac{1}{t}$  $(ii)$Multiplying equations $(i)$ and $(ii)$, we get:$(x-2y)(x+2y) = t 	imes \frac{1}{t}$$x^2 - 4y^2 = 1$This can be rewritten as:$\frac{x^2}{1} - \frac{y^2}{1/4} = 1$This represents a hyperbola with $a^2 = 1$ and $b^2 = \frac{1}{4}$.The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1/4}{1}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$.The centre is $(0,0)$ and the foci are $(\pm ae, 0) = \left(\pm 1 	imes \frac{\sqrt{5}}{2}, 0ight) = \left(\pm \frac{\sqrt{5}}{2}, 0ight)$.

For the variable $t,$ the locus of the points of intersection of lines $x-2y=t$ and $x+2y=\frac{1}{t}$ is

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