Find the measure of angle between the curves $x^2 - y^2 =3$ and $x^2 + y^2 -4x +3 = 0$
$x^2 -y^2 =3$ $\;\; \cdots$ (1)
$x^2 + y^2 -4x + 3 = 0$ $\;\; \cdots$ (2)
Adding equations (1) and (2) gives
$2x^2 -4x=0$
i.e. $x \left(x-2\right)=0$ $\implies$ $x=0$ or $x =2$
$\therefore$ $\;$ From equation (1),
when $x=0$, $y^2 = -3$
$\therefore$ $\;$ $x=0$ is not a valid answer.
when $x=2$, $y^2 = 1$ $\implies$ $y = \pm 1$
$\therefore$ $\;$ The two curves intersect at the points $\left(2,1\right)$ and $\left(2,-1\right)$
Differentiating equation (1) w.r.t x gives
$2x -2 y \dfrac{dy}{dx}=0$
i.e. $\dfrac{dy}{dx}=\dfrac{x}{y}$
$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(2,1\right)} = 2 = m_1 \left(\text{say}\right)$
and $\dfrac{dy}{dx}\bigg|_{\left(2,-1\right)}=-2 = m_3 \left(\text{say}\right)$
Differentiating equation (2) w.r.t x gives
$2x + 2y \dfrac{dy}{dx}-4=0$
i.e. $\dfrac{dy}{dx}= \dfrac{2-x}{y}$
$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(2,1\right)} = 0 = m_2 \left(\text{say}\right)$
and $\dfrac{dy}{dx}\bigg|_{\left(2,-1\right)}=0 = m_4 \left(\text{say}\right)$
Let $\alpha$ be the angle between the two tangents at the point $\left(2,1\right)$
Then $\tan \alpha = \left|\dfrac{m_1 - m_2}{1+m_1 m_2}\right|= \left|\dfrac{2-0}{1+2\times0}\right|=2$
$\implies$ $\alpha = \tan^{-1}\left(2\right)$
Let $\beta$ be the angle between the two tangents at the point $\left(2,-1\right)$
Then $\tan \beta = \left|\dfrac{m_3 - m_4}{1+m_3 m_4}\right|= \left|\dfrac{-2-0}{1+\left(-2\right)\times0}\right|=2$
$\implies$ $\beta = \tan^{-1}\left(2\right)$
Now, angle between the tangents of curves = angle between the curves
$\therefore$ Angle between the two curves $=\tan^{-1}\left(2\right)$