The angle of intersection of the curves

Question:

The angle of intersection of the curves $x y=a^{2}$ and $x^{2}-y^{2}=2 a^{2}$ is:

A. $0^{\circ}$

B. $45^{\circ}$

C. $90^{\circ}$

D. $30^{\circ}$

Solution:

Given that the curves $x y=a^{2}$ and $x^{2}-y^{2}=2 a^{2}$

Differentiating both of them w.r.t. $x$,

$x \frac{d y}{d x}+y=0$ and $2 x-2 y \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=\frac{-y}{x}$ and $\frac{d y}{d x}=\frac{x}{y}$

Let $\mathrm{m}_{1}=\frac{-\mathrm{y}}{\mathrm{x}}$ and $\mathrm{m}_{2}=\frac{\mathrm{x}}{\mathrm{y}}$

$m_{1} \times m_{2}=-1$

So, the angle between the curves is $90^{\circ}$.

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