Question

Find the angle of intersection of the curves:

$x^2+y^2=a^2\sqrt{2}$ and $x^2-y^2=a^2.$

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

Answer 1

The correct option is A $\frac{\pi }{4}$

$x^2+y^2=a^2\sqrt{2}$........(1), $x^2-y^2=a^2.$.............(2)
$\left\{\begin{array}{cc}Forpointsofintersection& 2x^2=a^2(\sqrt{2}+1)\\ & 2y^2=a^2(\sqrt{2}-1)\end{array}\right\}...\left(A\right)$

Now from $\left(A\right),4x^2{y}^2=a^4.1$
$\therefore 2xy=a^2.$ Also $y^2-x^2=-a^2.$....................(B)

Differentiating eq(1) and (2)
$2x+2y{\left(\frac{dy}{dx}\right)}_I=0,\therefore {\left(\frac{dy}{dx}\right)}_I=-\frac{x}{y}=m_1$
$2x-2y{\left(\frac{dy}{dx}\right)}_{II}=0,\therefore {\left(\frac{dy}{dx}\right)}_{II}=\frac{x}{y}=m_2.$
If $\theta$ be the angle between the curves, then
$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}=\frac{-2\left(x/y\right)}{1-x^2/y^3}=\frac{-2xy}{y^2-x^2}.$
From B
Hence $\tan \theta =\frac{-a^2}{-a^2}=1\therefore \theta =\frac{\pi }{4}.$