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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 :
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The correct answer is:
3
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),
$2 x+4 \frac{d y}{d x}=0$
$\Rightarrow \quad m_2=-1$
$\because \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{\frac{1}{2}+1}{1-\frac{1}{2}}\right|=\left|\frac{\frac{3}{2}}{\frac{1}{2}}\right|$
$\therefore \tan \theta=3$
$x y=2$ $\ldots$ (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),
$2 x+4 \frac{d y}{d x}=0$
$\Rightarrow \quad m_2=-1$
$\because \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{\frac{1}{2}+1}{1-\frac{1}{2}}\right|=\left|\frac{\frac{3}{2}}{\frac{1}{2}}\right|$
$\therefore \tan \theta=3$
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