Q70 / 200 · WBJEE 2013 · Mathematics
For the variable $t,$ the locus of the points of intersection of lines $x-2 y=t$ and $x+2 y=\frac{1}{t}$ is
- Athe straight line $x=y$
- Bthe circle with centre at the origin and radius 1
- Cthe ellipse with centre at the origin and one focus $\left(\frac{2}{\sqrt{5}}, 0\right)$
- Correct answerDthe hyperbola with centre at the origin and one focus $\left(\frac{\sqrt{5}}{2}, 0\right)$
Solution
Given equation of lines are
$$
x-2 y=t
$$
and $\quad x+2 y=\frac{1}{t}$
On multiplying Eqs. (i) and (ii) we get
$$
\begin{array}{r}
(x-2 y)(x+2 y)=t \times \frac{1}{t} \\
\Rightarrow \quad x^{2}-4 y^{2}=1
\end{array}
$$
$\Rightarrow \quad \frac{(x-0)^{2}}{1}-\frac{(y-0)^{2}}{(1 / 4)}=1$
which represent a hyperbola.
Here, $a^{2}=1$ and $b^{2}=\frac{1}{4}$
$\therefore$ Eccentricity $(e)=\sqrt{\frac{a^{2}+b^{2}}{a^{2}}}=\sqrt{\frac{1+\frac{1}{4}}{1}}=\frac{\sqrt{5}}{2}$
and focus $=(\pm a e, 0)=\left(\pm 1 \times \frac{\sqrt{5}}{2}, 0\right)=\left(\pm \frac{\sqrt{5}}{2}, 0\right)$
and centre $=(0,0)$
Hence, the correct answer is option: 'D'.