Application of Derivatives 1 Question 6
####6. If $\theta$ denotes the acute angle between the curves, $y=10-x^{2}$ and $y=2+x^{2}$ at a point of their intersection, then $|\tan \theta|$ is equal to
(2019 Main, 9 Jan I)
(a) $\frac{7}{17}$
(b) $\frac{8}{15}$
(c) $\frac{4}{9}$
(d) $\frac{8}{17}$
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Answer:
Correct Answer: 6. (b)
Solution:
- Key Idea Angle between two curves is the angle between the tangents to the curves at the point of intersection.
Given equation of curves are
and $\quad \begin{aligned} y & =10-x^{2} \\ y & =2+x^{2}\end{aligned}$
For point of intersection, consider
$ \begin{aligned} & & 10-x^{2} & =2+x^{2} \\ \Rightarrow & & 2 x^{2} & =8 \\ \Rightarrow & & x^{2} & =4 \\ \Rightarrow & & x & = \pm 2 \end{aligned} $
Clearly, when $x=2$, then $y=6$ (using Eq. (i)) and when $x=-2$, then $y=6$
Thus, the point of intersection are $(2,6)$ and $(-2,6)$.
Let $m_{1}$ be the slope of tangent to the curve (i) and $m_{2}$ be the slope of tangent to the curve (ii)
For curve (i) $\frac{d y}{d x}=-2 x$ and for curve (ii) $\frac{d y}{d x}=2 x$
$\therefore$ At $(2,6)$, slopes $m_{1}=-4$ and $m_{2}=4$, and in that case
$ |\tan \theta|=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|=\left|\frac{4+4}{1-16}\right|=\frac{8}{15} $
At $(-2,6)$, slopes $m_{1}=4$ and $m_{2}=-4$ and in that case
$ |\tan \theta|=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|=\left|\frac{-4-4}{1-16}\right|=\frac{8}{15} $
