If $ heta$ denotes the acute angle between the curves $y=10-x^2$ and $y=2+x^2$, at a point of the intersection, then $| an heta|$ is equal to

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If $	heta$ denotes the acute angle between the curves $y=10-x^2$ and $y=2+x^2$, at a point of the intersection, then $|	an 	heta|$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} & y=10-x^2 ...(i)\ & y=2+x^2...(ii) \end{aligned}$ Solving (i) and (ii), we get $y=6$ From (i), $6=10-x^2 \Rightarrow x= \pm 2$ Differentiating (i) w.r.t. $x$, we get $\frac{\mathrm{d} y}{\mathrm{~d} x}=-2 x \Rightarrow\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}ight)_{(-2,6)}=4$ Differentiating (ii) w.r.t. $x$, we get $\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x \Rightarrow\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}ight)_{(-2,6)}=-4 \ 	herefore \quad & 	an 	heta=\left|\frac{4-(-4)}{1+4(-4)}ight|=\left|\frac{8}{-15}ight| \Rightarrow|	an 	heta|=\frac{8}{15} \end{aligned}$

If $\theta$ denotes the acute angle between the curves $y=10-x^2$ and $y=2+x^2$, at a point of the…

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