Find the equation of the tangent to the curve $y=\sqrt{3x-2}$ which is parallel to the line $4x-2y+5=0$
Differentiating the given curve w.r.t x gives
$\dfrac{dy}{dx}=\dfrac{3}{2\sqrt{3x-2}}$ $\;\; \cdots$ (1)
Slope of the line $4x-2y+5=0$ is 2 $\cdots$ (2)
Since the tangent to the curve is parallel to the given line
$\implies$ Slope of tangent $=$ Slope of the line
$\therefore$ $\;$ From equations (1) and (2) we have
$\dfrac{3}{2\sqrt{3x-2}} = 2$
i.e. $3x-2 = \dfrac{9}{16}$ $\implies$ $x = \dfrac{41}{48}$
Substituting the value of x in the equation of curve gives
$y = \sqrt{\dfrac{123}{48}-2} = \sqrt{\dfrac{123-96}{48}}=\sqrt{\dfrac{27}{48}} = \dfrac{3}{4}$
$\therefore$ $\;$ From equation (1), slope of tangent at $\left(\dfrac{41}{48}, \dfrac{3}{4}\right)$ is
$\dfrac{dy}{dx} \bigg|_{\left(\frac{41}{48}, \frac{3}{4}\right)} = \dfrac{3}{2\sqrt{\left(3\times \dfrac{41}{48}\right)-2}} = \dfrac{3}{2 \sqrt{\dfrac{9}{16}}}=2$
$\therefore$ $\;$ Equation of tangent to the given curve parallel to the line $4x-2y+5=0$ is
$y - \dfrac{3}{4} = 2 \left(x - \dfrac{41}{48}\right)$
i.e. $4y-3=8 \left(\dfrac{48x-41}{48}\right)$
i.e. $24y - 18 = 48x - 41$
i.e. $48x - 24y -23=0$