Question

The angle between the curves $x^2+y^2=\sqrt{2}{a}^2$ and $x^2-y^2=a^2$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$\frac{\pi }{4}$

$y^{\prime }=\frac{-x}{y}=m_1$
$y^{\prime }=\frac{x}{y}=m_2$
$2x^2=\sqrt{2}{a}^2+a^2$
$x=a\left(\frac{\sqrt{2}+1}{2}\right){}^1{/}_2$
$\left(\frac{\sqrt{2}{a}^2+a^2}{2}\right)-y^2=a^2$
$y^2=(\sqrt{2}-1)a^2$
$y=\left(\frac{\sqrt{2}-1}{2}\right)a^{{}_2^1}$
$m_1={\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)}_2^1$
$m_2=-(\sqrt{2}+1)$
$tan\theta =\left|\frac{(\sqrt{2}+1+\sqrt{2}-1)}{1-(\sqrt{2}+1{)}^2}\right|$
$=1$
$\therefore \theta =\pi /4$