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

Question

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

Dr. Abubacker · Accepted Answer

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

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

Explanation

Reviewed by Dr. Abubacker

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