Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. $3$

Answer 1

The correct option is D

$3$

Explanation for the correct option:

Step 1: Find the point of intersection of the curves $xy=2$ and $x^2+4y=0$:

$$\begin{gathered} x^2+4y=0 \\ \Rightarrow 4y=-x^2 \\ \Rightarrow y=\frac{-x^2}{4} \end{gathered}$$

Substitute value of $y$ in $xy=2$, then

$$\begin{gathered} xy=2 \\ \Rightarrow x\left(\frac{-x^2}{4}\right)=2 \\ \Rightarrow x^3=-8 \\ \Rightarrow x=-2 \end{gathered}$$

Substitute value of $x$ in $xy=2$,

$$\begin{gathered} xy=2 \\ \Rightarrow -2y=2 \\ \Rightarrow y=-1 \end{gathered}$$

So, the point of intersection is $(-2,-1)$.

Step 2: Finding the slope:

Differentiate $xy=2$ with respect to $x$ both sides

$$\begin{gathered} x\frac{dy}{dx}+y=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-y}{x} \end{gathered}$$

Let $m_1$ denotes the slope of $xy=2$ .

Then,

$\begin{array}{rcl}{m}_1& =& \frac{dy}{dx}\\ & =& \frac{-y}{x}\\ & =& \frac{-1}{2}\end{array}$

Differentiate $x^2+4y=0$ with respect to $x$ both sides

$$\begin{gathered} 2x+4\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-2x}{4} \\ \Rightarrow \frac{dy}{dx}=\frac{-x}{2} \end{gathered}$$

Let $m_2$ denotes the slope of $x^2+4y=0$

Then,

$\begin{array}{rcl}{m}_2& =& \frac{dy}{dx}\\ & =& \frac{-x}{2}\\ & =& 1\end{array}$

Step 3: Find $tan\theta$:

$$\begin{gathered} \tan \theta =\left|\frac{{\displaystyle \frac{-1}{2}}-1}{1-{\displaystyle \frac{1}{2}}}\right|\left[\because tan\theta =\left|\frac{(m_1–m_2)}{1+m_1{m}_2}\right|\right] \\ \Rightarrow \tan \theta =\left|-3\right| \\ \Rightarrow \tan \theta =3 \end{gathered}$$

Hence, the correct option is D.