Question

Find the angle of intersection of curves $y=4-x^2$ and $y=x^2$

• A. ${\tan }^{-1}\left(\frac{4\sqrt{2}}{5}\right)$
• B. ${\tan }^{-1}\left(\frac{2\sqrt{2}}{7}\right)$
• C. ${\tan }^{-1}\left(\frac{4\sqrt{2}}{7}\right)$
• D. None of these

Answer 1

Answer: (C)

${\tan }^{-1}\left(\frac{4\sqrt{2}}{7}\right)$

Let
$y_1=4-x^2$ and $y_2=x^2$

Thus

$4-x^2=x^2$
$2x^2=4$
$x^2=2$
$x=\pm \sqrt{2}$

$y=2$

Hence the curves intersect at

$(-\sqrt{2},2)$ and $(\sqrt{2},2)$

Now slope of the curves at $(\sqrt{2},2)$ is

$y_1^{\prime }{|}_{x=\sqrt{2}}=-2x{|}_{\sqrt{2}}$

$=-2\sqrt{2}$

$y_2^{\prime }{|}_{\sqrt{2}}=2x{|}_{\sqrt{2}}$

$=2\sqrt{2}$

Hence

$\tan \theta =|\frac{y_2^{\prime }-y_1^{\prime }}{1+y_1^{\prime }{y}_2^{\prime }}|$

$=|\frac{4\sqrt{2}}{1-8}|$

$\tan \theta =\frac{4\sqrt{2}}{7}$

$\theta ={\tan }^{-1}\left(\frac{4\sqrt{2}}{7}\right)$