Question
If $$q$$ denotes the acute angle between the curves, $$y = 10 - {x^2}$$ and $$y = 2 + {x^2}$$ at a point of their intersection, then $$\left| {\tan \theta } \right|$$ is equal to:
A.
$$\frac{4}{9}$$
B.
$$\frac{8}{{15}}$$
C.
$$\frac{7}{{17}}$$
D.
$$\frac{8}{{17}}$$
Answer :
$$\frac{8}{{15}}$$
Solution :
Since, the equation of curves are
$$y = 10 - {x^2}\,........\left( 1 \right)\,\,\,y = 2 + {x^2}\,.......\left( 2 \right)$$
Adding equation (1) and (2), we get
$$2y = 12 \Rightarrow y = 6$$
Then, from equation (1)
$$x = \pm 2$$
Differentiate equation (1) with respect to $$x$$
$$\frac{{dy}}{{dx}} = - 2x \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {2,6} \right)}} = - 4\,{\text{and}}\,{\left( {\frac{{dy}}{{dx}}} \right)_{\left( { - 2,6} \right)}} = 4$$
Differentiate equation (2) with respect to $$x$$
$$\eqalign{
& \frac{{dy}}{{dx}} = 2x \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {2,6} \right)}} = 4{\text{ and }}{\left( {\frac{{dy}}{{dx}}} \right)_{\left( { - 2,6} \right)}} = - 4 \cr
& \operatorname{At} \,\left( {2,6} \right)\tan \theta = \left( {\frac{{\left( { - 4} \right) - \left( 4 \right)}}{{1 + \left( { - 4} \right) \times \left( 4 \right)}}} \right) = \frac{{ - 8}}{{15}} \cr
& \operatorname{At} \,\left( { - 2,6} \right),\tan \theta = \frac{{\left( 4 \right) - \left( { - 4} \right)}}{{1 + \left( 4 \right)\left( { - 4} \right)}} = \frac{8}{{ - 15}} \cr
& \therefore \,\left| {\tan \theta } \right| = \frac{8}{{15}} \cr} $$