Question

The coordinates of the point on the curve y = 2 + $\sqrt{4x+1}$ where tangent has slope $\frac{2}{5}$ are _________________.

Answer 1

Let (h, k) be the point on the curve $y=2+\sqrt{4x+1}$ where tangent has slope $\frac{2}{5}$.

$y=2+\sqrt{4x+1}$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=0+\frac{1}{2\sqrt{4x+1}}\times 4$

$\Rightarrow \frac{dy}{dx}=\frac{2}{\sqrt{4x+1}}$

∴ Slope of tangent at (h, k) = ${\left(\frac{dy}{dx}\right)}_{\left(h,k\right)}=\frac{2}{5}$ (Given)

$\Rightarrow \frac{2}{\sqrt{4h+1}}=\frac{2}{5}$

$\Rightarrow \sqrt{4h+1}=5$

$\Rightarrow 4h+1=25$

$\Rightarrow 4h=24$

$\Rightarrow h=6$

Now, (h, k) lies on the given curve $y=2+\sqrt{4x+1}$.

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

Putting h = 6 in (1), we get

$k=2+\sqrt{4\times 6+1}$

$\Rightarrow k=2+\sqrt{25}$

$\Rightarrow k=2+5=7$

So, the coordinates of the required point are (6, 7).

Thus, the coordinates of the point on the curve $y=2+\sqrt{4x+1}$ where tangent has slope $\frac{2}{5}$ are (6, 7).


The coordinates of the point on the curve y = 2 + $\sqrt{4x+1}$ where tangent has slope $\frac{2}{5}$ are ___(6, 7)___.