Question

Find the equation of the tangent to the curve $y=\sqrt{3x-2}$ which is parallel to the 4x − 2y + 5 = 0.

Answer 1

Slope of the given line is 2
$$\begin{gathered} \text{Let}\left(x_1,y_1\right)\text{be the point where the tangent is drawn to the curve}y=\sqrt{3x-2} \\ \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},y_1=\sqrt{3x_1-2}...\left(1\right) \\ \mathrm{Now},y=\sqrt{3x-2} \\ \Rightarrow \frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}} \\ \text{Slope of tangent at}\left(x_1,y_1\right)\text{=}\frac{3}{2\sqrt{3x_1-2}} \\ \text{Given that} \\ \text{Slope of}\mathrm{tangent}\text{= slope of the given line} \\ \Rightarrow \frac{3}{2\sqrt{3x_1-2}}=2 \\ \Rightarrow 3=4\sqrt{3x_1-2} \\ \Rightarrow 9=16\left(3x_1-2\right) \\ \Rightarrow \frac{9}{16}=3x_1-2 \\ \Rightarrow 3x_1=\frac{9}{16}+2=\frac{9+32}{16}=\frac{41}{16} \\ \Rightarrow x_1=\frac{41}{48} \\ \mathrm{Now},y_1=\sqrt{\frac{123}{48}-2}=\sqrt{\frac{27}{48}}=\sqrt{\frac{9}{16}}=\frac{3}{4}\left[\mathrm{From}\text{(1)}\right] \\ \therefore \left(x_1,y_1\right)=\left(\frac{41}{48},\frac{3}{4}\right) \\ \text{Equation of}\mathrm{tangent}\text{is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-\frac{3}{4}=2\left(x-\frac{41}{48}\right) \\ \Rightarrow \frac{4y-3}{4}=2\left(\frac{48x-41}{48}\right) \\ \Rightarrow 24y-18=48x-41 \\ \Rightarrow 48x-24y-23=0 \end{gathered}$$