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

Toolbox:
• Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
• Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1) Step 1: Let the point of contact of the tangent line parallel to the given line be P(x_1,y_1). The equation of the curve is y=\sqrt{3x-2} Now differentiating w.r.t x we get, \large\frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}} Hence the slope of the tangent at (x_1,y_1) is \large\frac{dy}{dx}_{(x_1,y_1)}=\frac{3}{2\sqrt{3x_1-2}} Step 2: Since the tangent is at (x_1,y_1) is parallel to the line 4x-2y+5=0 Therefore \large\frac{dy}{dx}_{(x_1,y_1)}$$=$slope of the line $4x-2y+5=0$
Slope of the line is $-\bigg(\large\frac{coeff\;of\;x}{coeff\;of\;y}\bigg)$
$\Rightarrow -\big(\large\frac{-4}{2}\big)$$=2 Therefore \large\frac{3}{2\sqrt{3x_1-2}}$$=2$
$\Rightarrow 3=4\sqrt{3x_1-2}$
Step 3:
Squaring on both sides we get,
$9=16(3x_1-2)$
Therefore $3x_1=\large\frac{9}{16}$$+2 \qquad\qquad\quad=\large\frac{41}{16} Therefore x_1=\large\frac{41}{48} Since (x_1,y_1) lies on y=\sqrt{3x-2}, y_1=\sqrt{3x_1-2} \quad=\sqrt{3\times \large\frac{41}{48}-\normalsize 2} \Rightarrow y_1=\large\frac{3}{4} Hence the point of contact is \big(\large\frac{41}{48},\frac{3}{4}\big) Step 4: Hence the required equation of the tangent is y-\large\frac{3}{4}$$=2(x-\large\frac{41}{48})$
$\Rightarrow \large\frac{4y-3}{4}=\frac{48x-41}{24}$
$\Rightarrow 6(4y-3)=48x-41$
Therefore $48x-24y=23$
This is the required equation of the tangent.