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 $90^{\circ}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.