Question

The cosine of the acute angle between the curves $y=\left|x^{2} - 1\right|$ and $y=\left|x^{2} - 3\right|$ at their points of intersection is

• A. $\frac{1}{3}$
• B. $\frac{7}{9}$
• C. $\frac{11}{9 \sqrt{2}}$
• D. $\frac{2}{7}$

Answer 1

Answer: (B)

For points of intersection $P\&Q$ , solving the curves $\left|x^{2} - 1\right|=\left|x^{2} - 3\right|,$ we get,
$\Rightarrow x^{2}-1=3-x^{2}$
$\Rightarrow x=\pm\sqrt{2}\Rightarrow P\left(\sqrt{2} , 1\right)\&Q\left(- \sqrt{2} , 1\right)$
For the slope of tangents at $x=\sqrt{2}$
$C_{1}$ is $y=x^{2}-1$
$\frac{d y}{d x}=2x$
$\Rightarrow m_{1}=2\sqrt{2}$
And $C_{2}$ is $y=3-x^{2}$
$\frac{d y}{d x}=-2x$
$\Rightarrow m_{2}=-2\sqrt{2}$
$\therefore tan \theta =\left|\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}\right|$
$=\left|\frac{4 \sqrt{2}}{1 - 8}\right|=\frac{4 \sqrt{2}}{7}$
$\therefore cos \theta =\frac{7}{9}$