First slide

Tangents and normals

Question

The slope of the normal to the curve $x^{2} + y^{2} = 4$ at $(\sqrt{2}, \sqrt{2})$ is

Easy

A

0
B

1
C

$- 1$
D

$\frac{1}{\sqrt{2}}$

Solution

$\frac{d y}{d x} = \frac{- x}{y}$
$m = {(\frac{d y}{d x})}_{(\sqrt{2} \sqrt{2})} = - 1$

Slope of the normal = 1

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