Q71 / 120 · VITEEE 2014 · Mathematics
The angle between lines joining the origin to the point of intersection of the line $\sqrt{3} x+y=2$ and the curve $y^{2}-x^{2}=4$ is
- A$\tan ^{-1} \frac{2}{\sqrt{3}}$
- B$\frac{\pi}{6}$
- Correct answerC$\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
- D$\frac{\pi}{2}$
Solution
On homogenising $y^{2}-x^{2}=4$ with the help of the line $\sqrt{3} x+y=2$, we get
$\begin{aligned}
& y^{2}-x^{2}=4 \frac{(\sqrt{3} x+y)^{2}}{4} \\
\Rightarrow & y^{2}-x^{2}=3 x^{2}+y^{2}+2 \sqrt{3} x y \\
\Rightarrow & 4 x^{2}+2 \sqrt{3} x y=0
\end{aligned}$
On comparing with $a x^{2}+2 h x y+b y^{2}=0$, we get
$\mathrm{a}=4, \mathrm{~h}=\sqrt{3} \text { and } \mathrm{b}=0$
We know that,
$\tan \theta=2 \frac{\sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}=\frac{2 \cdot \sqrt{3-0}}{4+0}$
$\therefore \quad$ The angle between the lines is
$\theta=\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
Hence, the correct answer is option: 'C'.