Question

Find the point at where the tangent to the curve y=$\sqrt{4x-3}-1$ has its slope $\frac{2}{3}$

Answer 1

Equation of the curve : $y=\sqrt{4x-3}-1$

To find the slope of the tangent, we need to differentiate the equation of the curve,

$\frac{dy}{dx}=\frac{1}{2}\times \frac{1}{\sqrt{4x-3}}\times 4=\frac{2}{\sqrt{4x-3}}$

Given slope of tangent is $\frac{2}{3}$

Thus, $\frac{2}{\sqrt{4x-3}}=\frac{2}{3}\Rightarrow \sqrt{4x-3}=3\Rightarrow 4x-3=9\Rightarrow 4x=12\Rightarrow x=3$

Putting the value of x in equation of the curve, we get,

$y=\sqrt{4x-3}-1\Rightarrow y=\sqrt{4\times 3-3}-1=\sqrt{9}-1=3-1=2$

Thus the point at which tangent to the curve $y=\sqrt{4x-3}-1$ has slope $\frac{2}{3}$ is $(3,2)$