If $ heta$ is the angle between the curves $xy = 2$ and $x^{2} + 4y = 0$, then $ an heta$ is equal to

Question

If $	heta$ is the angle between the curves $xy = 2$ and $x^{2} + 4y = 0$, then $	an	heta$ is equal to

Dr. Abubacker · Accepted Answer

(d)
The point of intersection of given curves are (2, -1)
On differentiating the given curves respectively
$x\frac{dy}{dx} + y = 0\ \ \ \  \Rightarrow \ \ \frac{dy}{dx} = - \frac{y}{x}$
$\Rightarrow \ \ \ \ m_{1} = \left( \frac{dy}{dx} ight)_{(2,\  - 1)} = \frac{1}{2}$
And $2x + 4\frac{dy}{dx} = 0$
$\Rightarrow \ \ \ \ \ \ \ m_{2} = \left( \frac{dy}{dx} ight)_{(2,\  - 1)} = - 1$
$\because\ \ \ \ \ \ 	an	heta = \left| \frac{\frac{1}{2} + 1}{1 - \frac{1}{2}} ight| = \left| \frac{\frac{3}{2}}{\frac{1}{2}} ight| = 3$

Q. If $\theta$ is the angle between the curves $xy = 2$ and $x^{2} + 4y = 0$, then $\tan\theta$ is equal to

Explanation

Reviewed by Dr. Abubacker

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