Question

The equation of the tangent to the curve y = x + $\frac{4}{x^2}$, that is parallel to x-axis, is ________________.

Answer 1

The equation of the given curve is $y=x+\frac{4}{x^2}$.

Let (h, k) be the point of contact of tangent with the curve.

$\therefore k=h+\frac{4}{h^2}$ .....(1)

$y=x+\frac{4}{x^2}$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=1-\frac{8}{x^3}$

∴ Slope of tangent at (h, k) $={\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=1-\frac{8}{h^3}$

It is given that, the tangent to the given curve is parallel to the x-axis.

∴ Slope of tangent = 0

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=0$

$\Rightarrow 1-\frac{8}{h^3}=0$

$\Rightarrow h^3=8$

$\Rightarrow h=2$

Putting h = 2 in (1), we get

$k=2+\frac{4}{{\left(2\right)}^2}=2+1=3$

Thus, the point of contact of tangent with the curve is (2, 3).

So, the equation of the tangent at (2, 3) is

$y-3={\left(\frac{dy}{dx}\right)}_{\left(2,3\right)}\left(x-2\right)$

$\Rightarrow y-3=0\left[{\left(\frac{dy}{dx}\right)}_{\left(2,3\right)}=0\right]$

$\Rightarrow y=3$

Thus, the equation of tangent to the given curve $y=x+\frac{4}{x^2}$, that is parallel to x-axis, is y = 3.


The equation of the tangent to the curve y = x + $\frac{4}{x^2}$, that is parallel to x-axis, is ___y = 3___.