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The equation of normal to the curve y = tanx at (0, 0) is ________.

Answers (1)

Given curve is y = tanx

Apply first derivative and get

\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}(\tan \mathrm{x})}{\mathrm{dx}}$ \\$\Rightarrow \frac{d y}{d x}=\sec ^{2} x$
It is the slope of tangent
Substitute (0,0) in slope and get
\\\left(\frac{d y}{d x}\right)_{(0,0)}=\sec ^{2} x$ \\$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0,0)}=\sec ^{2} 0=1$
Hence, -1 is the slope of the normal to the curve at (0,0)
Hence the equation is
\\$y-0=-1(x-0)$ \\$\Rightarrow y=-x$ \\$or $x+y=0$

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