Question

The equation of the normal to the curve y = tan x at (0, 0) is ______________.

Answer 1

The given curve is

y = tanx

Differentiating both sides with respect to x, we have

$\frac{dy}{dx}={\sec }^{2}x$

∴ Slope of tangent at (0, 0) = ${\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}={\sec }^{2}0={\left(1\right)}^2=1$

⇒ Slope of normal at (0, 0) = $-\frac{1}{{\left({\displaystyle \frac{dy}{dx}}\right)}_{\left(0,0\right)}}=-\frac{1}{1}=-1$

So, the equation of normal at (0, 0) is

$y-0=-\frac{1}{{\left({\displaystyle \frac{dy}{dx}}\right)}_{\left(0,0\right)}}\left(x-0\right)$

$\Rightarrow y=-1\times x$

$\Rightarrow x+y=0$

Thus, the equation of the normal to the given curve at (0, 0) is x + y = 0.


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