Question

The tangent to the curve y = e^2x at (0, 1) cuts x-axis at the point __________________.

Answer 1

The given curve is

y = e^2x

Differentiating both sides with respect to x, we get

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

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

So, the equation of tangent at (0, 1) is

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

$\Rightarrow y-1=2x$ [Using (1)]

$\Rightarrow 2x-y+1=0$ .....(2)

This tangent cuts the x-axis where y = 0.

Putting y = 0 in (2), we get

$2x-0+1=0$

$\Rightarrow x=-\frac{1}{2}$

So, the coordinates of the required point are $\left(-\frac{1}{2},0\right)$.

Thus, the tangent to the given curve y = e^2x at (0, 1) cuts the x-axis at $\left(-\frac{1}{2},0\right)$.


The tangent to the curve y = e^2x at (0, 1) cuts x-axis at the point $\underline{\left(-\frac{1}{2},0\right)}$.