Question

Consider the curves $y=x^2+2$ and$y=10-x^2$. Let $\theta$ be the angle between both the curves at the point of intersection, then find $\left|\tan \theta \right|$

• A. $\frac{8}{15}$
• B. $\frac{5}{17}$
• C. $\frac{3}{17}$
• D. $\frac{8}{17}$

Answer 1

The correct option is A

$\frac{8}{15}$

Step$1$. To find the point of intersection:

Putting $y=x^2+2$ in $y=10-x^2$, we get

$$\begin{gathered} x^2+2=10-x^2 \\ \Rightarrow 2x^2=8 \\ \Rightarrow x=\pm 2 \\ \Rightarrow y=2^2+2=6 \end{gathered}$$

So, the point of intersection is $\left(\pm 2,6\right)$

Step$2$. To find $\left|\tan \theta \right|$:

$y=x^2+2$and $y=10-x^2$

$\Rightarrow \frac{dy}{dx}=2x$and $\frac{dy}{dx}=-2x$

$\Rightarrow \frac{dy}{dx}$at $\left(\pm 2,6\right)$$=\pm 4=m_1$and $\frac{dy}{dx}$at $\left(\pm 2,6\right)$$=\pm 4=m_2$

$$\begin{gathered} \Rightarrow \left|\tan \theta \right|=\left|\frac{4-(-4)}{1-16}\right|\mathbf{[}\mathbf{}\left|\tan \theta \right|\mathbf{}\mathbf{=}\mathbf{}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|\mathbf{]} \\ \Rightarrow \left|\tan \theta \right|=\left|\frac{8}{1-16}\right| \\ \Rightarrow \left|\tan \theta \right|=\left|\frac{8}{15}\right| \end{gathered}$$

Hence, the correct option is (A).