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

$$\text{ Let }(x_1,y_1)\text{ be the point where the tangent is drawn to the curvey }=\sqrt{3x-2}$$

$$\text{ Since, the point lies on the curve }.$$

$$\text{ Hence },y_1=\sqrt{3x_1-2}...\left(1\right)$$

$$\text{ Now },y=\sqrt{3x-2}$$

$$\Rightarrow \frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}$$

$$\text{ Slope of tangent at}(x_1,y_1)=\frac{3}{2\sqrt{3x_1-2}}$$

$$\text{ Given that }$$

$$\text{ Slope of tangent = slope of the given line }$$

$$\Rightarrow \frac{3}{2\sqrt{3x_1-2}}=2$$

$$\Rightarrow 3=4\sqrt{3x_1-2}$$

$$\Rightarrow 9=16(3x_1-2)$$

$$\Rightarrow \frac{9}{16}=3x_1-2$$

$$\Rightarrow 3x_1=\frac{9}{16}+2=\frac{9+32}{16}=\frac{41}{16}$$

$$\Rightarrow x_1=\frac{41}{48}$$

$$\text{ Now,}{y}_1=\sqrt{\frac{123}{48}-2}=\sqrt{\frac{27}{48}}=\sqrt{\frac{9}{16}}=\frac{3}{4}..............[\text{ From }(1)]$$

$$\therefore (x_1,y_1)=(\frac{41}{48},\frac{3}{4})$$

$$\text{ Equation of tangent is },$$

$$y-y_1=m(x-x_1)$$

$$\Rightarrow y-\frac{3}{4}=2(x-\frac{41}{48})$$

$$\Rightarrow \frac{4y-3}{4}=2\left(\frac{48x-41}{48}\right)$$

$$\Rightarrow 24y-18=48x-41$$

$$\Rightarrow 48x-24y-23=0$$