Q108 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

An angle between the curves $x^2-y^2=4$ and $x^2+y^2$ $=4 \sqrt{2}$ is

  • A
    $\frac{\pi}{6}$
  • Correct answer
    B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{3}$
  • D
    $\frac{\pi}{2}$

Solution

$x^2-y^2=4$ ...(i) $x^2+y^2=4 \sqrt{2}$ ...(ii) Adding both equations, $\begin{aligned} & x^2=2(1+\sqrt{2}) \Rightarrow x=\sqrt{2} \sqrt{\sqrt{2}+1} \\ & \therefore \quad y^2=2(\sqrt{2}-1) \Rightarrow y=\sqrt{2} \sqrt{\sqrt{2}-1}\end{aligned}$ Differentiating both (i) and (ii) $\begin{aligned} & 2 x-2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{x}{y}=m_1 \\ & \text { and } 2 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-x}{y}=m_2 \\ & \tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\left|\frac{\frac{x}{y}+\frac{x}{y}}{1-\frac{x^2}{y^2}}\right| \\ & \quad=\left|2 \frac{x}{y} \times \frac{y^2}{y^2-x^2}\right|=\left|\frac{2 x y}{y^2-x^2}\right|=\left|\frac{4}{-4}\right|=1 . \\ & \therefore \quad \theta=\frac{\pi}{4} .\end{aligned}$

Hence, the correct answer is option: 'B'.

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