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Q.
An angle between the curves $x^{2}-y^{2}=4$ and $x^{2}+y^{2}=4 \sqrt{2}$ is
TS EAMCET 2018
Solution:
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'
$2 x-2 y \frac{d y}{d x}=0$ $ \frac{d y}{d x}=\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'
$2 x+2 y \frac{d y}{d x}=0 $ $\frac{d y}{d x}=-\frac{x}{y}$
At point $\left(x_{0}, y_{0}\right)$
$m_{2}=-\frac{x_{0}}{y_{0}}$
Angle between tangents $\theta=\tan ^{-1}\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
$\theta=\tan ^{-1}\left|\frac{\frac{x_{0}}{y_{0}}+\frac{x_{0}}{y_{0}}}{1+\frac{x_{0}}{y_{0}}\left(-\frac{x_{0}}{y_{0}}\right)}\right|$
$\Rightarrow \theta=\tan ^{-1}\left|\frac{2 \frac{x_{0}}{y_{0}}}{\frac{y_{0}^{2}-x_{0}^{2}}{y_{0}^{2}}}\right|$
$\Rightarrow \, \theta=\tan ^{-1}\left(\frac{2 x_{0} y_{0}}{y_{0}^{2}-x_{0}^{2}}\right)=\tan ^{-1}\left(\frac{2 x_{0} y_{0}}{4}\right)$
[given, $y^{2}-x^{2}=4$ satisfied point $\left.\left(x_{0}, y_{0}\right)\right]$
$=\tan ^{-1}\left(\frac{x_{0} y_{0}}{2}\right)\,\,\,\,\,\,\dots(i)$
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) $
$\Rightarrow \,\theta=\frac{\pi}{4}$