ISI B.Math Entrance (7)

We need to find the slope of the 2 lines 1st.

So $xy=1$

Differentiating w.r.t $x$ , we get

$y + x\dfrac{dy}{dx}= 0$ $\implies$ $\dfrac{dy}{dx} = \dfrac{-y}{x} .... (i)$

Now differentiating $x^2-y^2=1$ w.r.t $x$ , we get

$\implies 2x - 2y \dfrac{dy}{dx} = 0$ $\implies \dfrac{dy}{dx} = \dfrac{x}{y}......(ii)$

So $(i) \times (ii)$ $\implies \dfrac{-y}{x} \times \dfrac{x}{y}$ $\implies -1$

So They are perpendicular to each other, or the product of their slopes in $-1$

So angle between the curves at the point of intersection will be $\dfrac{\pi}{2}$

Using the grad function on both curves: $\nabla (xy) = (y,x) \quad \nabla (x^2 -y^2) = 2(x,-y)$ so these are the tangents to the curves $\ xy = 1$ and $\ x^2 -y^2 = 1$ respectively (at points (x,y)). Dotting these gives: $\ (y,x) \cdot (x,-y) = yx - xy = 0$ So the curves intersect at right angles, that is, at $\frac{\pi}{2}$ radians.