Question

An angle between the curves $x^2-y^2=4$ and $x^2+y^2=4\sqrt{2}$ is

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

Answer 1

Answer: (B)

Equation of given curves, $x^2-y^2=4$ and
$x^2+y^2=4\sqrt{2}$
Let $\left(x_0,y_0\right)$ be point of intersection.
$\therefore$ Angle between curves $=$ Angle between tangent at point of intersection.
Now, $x^2-y^2=4$
Differentiate w.r.t. 'x'
$2x-2y\frac{dy}{dx}=0$ $\frac{dy}{dx}=\frac{x}{y}$
At point $\left(x_0,y_0\right)$
$m_1=\frac{x_0}{y_0}$ and curve $x^2+y^2=4\sqrt{2}$
Differentiate w.r.t. 'x'
$2x+2y\frac{dy}{dx}=0$ $\frac{dy}{dx}=-\frac{x}{y}$
At point $\left(x_0,y_0\right)$
$m_2=-\frac{x_0}{y_0}$
Angle between tangents $\theta ={\tan }^{-1}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
$\theta ={\tan }^{-1}\left|\frac{\frac{x_0}{y_0}+\frac{x_0}{y_0}}{1+\frac{x_0}{y_0}\left(-\frac{x_0}{y_0}\right)}\right|$
$\Rightarrow \theta ={\tan }^{-1}\left|\frac{2\frac{x_0}{y_0}}{\frac{y_0^2-x_0^2}{y_0^2}}\right|$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{2x_0{y}_0}{y_0^2-x_0^2}\right)={\tan }^{-1}\left(\frac{2x_0{y}_0}{4}\right)$
[given, $y^2-x^2=4$ satisfied point $\left(x_0,y_0\right)]$
$={\tan }^{-1}\left(\frac{x_0{y}_0}{2}\right)\dots (i)$
Point $\left(x_0,y_0\right)$ lie on curves, so
$x_0^2-y_0^2=4$ and $x_0^2+y_0^2=4\sqrt{2}$
$\Rightarrow x_0^2=2(1+\sqrt{2})\text{ and }{y}_0^2=2(\sqrt{2}-1)$
So, $x_0^2{y}_0^2=4(\sqrt{2}+1)(\sqrt{2}-1)=4(2-1)=4$
$x_0{y}_0=2$
Put in Eq. (i),
$\theta ={\tan }^{-1}\left(\frac{2}{2}\right)$
$\theta ={\tan }^{-1}(1)$
$\Rightarrow \theta =\frac{\pi }{4}$