Question

Find the angle between the curves X^2-Y^2=1 and XY=√2 at (√2, 1)

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{S_{1}:{x}^{2}-{y}^{2}=1\,\,\,\,\,\&\,\,\,\,\,S_{2}:xy=\sqrt{2}}$

$\tt{\implies\,2x-2y\cdot\dfrac{dy}{dx}=0\,\,\,\,\,\&\,\,\,\,\,y+x\cdot\dfrac{dy}{dx}=0}$

$\tt{\implies\,2y\cdot\dfrac{dy}{dx}=2x\,\,\,\,\,\&\,\,\,\,\,x\cdot\dfrac{dy}{dx}=-y}$

$\tt{\implies\,\dfrac{dy}{dx}=\dfrac{x}{y}\,\,\,\,\,\&\,\,\,\,\,\dfrac{dy}{dx}=-\dfrac{y}{x}}$

$\tt{\implies\,\dfrac{dy}{dx}\bigg|_{\left(\sqrt{2},1\right)}=\dfrac{\sqrt{2}}{1}\,\,\,\,\,\&\,\,\,\,\,\dfrac{dy}{dx}\bigg|_{\left(\sqrt{2},1\right)}=-\dfrac{1}{\sqrt{2}}}$

$\tt{\implies\,m_{1}=\sqrt{2}\,\,\,\,\,\&\,\,\,\,\,m_{2}=-\dfrac{1}{\sqrt{2}}}$

Since  $\tt{m_{1}\cdot\,m_{2}=-1}$

Hence, at  $\left(\sqrt{2},1\right)$, the angle between the given curves is 90°.