Q107 / 160 · AP EAMCET 2006 · Mathematics

If $\theta$ is the angle between the curves $x y=2$ and $x^2+4 y=0$ and $x^2+4 y=0$, then $\tan \theta$ is equal to :

  • A
    1
  • B
    -1
  • C
    2
  • Correct answer
    D
    3

Solution

The equation of curves are $x y=2$ $\ldots$ (i) and $\quad x^2+4 y=0$ $\ldots$ (ii) On solving Eqs. (i) and (ii), we get Point of intersection which is $(2,-1)$. From Eq. (i) $x \frac{d y}{d x}+y=0$ $\Rightarrow \quad \frac{d y}{d x}=-\frac{y}{x}$ $m_1=\left(\frac{d y}{d x}\right)_{(2,-1)}=\frac{1}{2}$ and From Eq. (ii), $2 x+4 \frac{d y}{d x}=0$ $\Rightarrow \quad m_2=-1$ $\because \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{\frac{1}{2}+1}{1-\frac{1}{2}}\right|=\left|\frac{\frac{3}{2}}{\frac{1}{2}}\right|$ $\therefore \tan \theta=3$

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

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