# Find the equation of the tangent to the curve

Question:

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

Solution:

The equation of the given curve is $y=\sqrt{3 x-2}$.

The slope of the tangent to the given curve at any point (xy) is given by,

$\frac{d y}{d x}=\frac{3}{2 \sqrt{3 x-2}}$

The equation of the given line is $4 x-2 y+5=0$.

$4 x-2 y+5=0 \Rightarrow y=2 x+\frac{5}{2}($ which is of the form $y=m x+c)$

∴Slope of the line = 2

Now, the tangent to the given curve is parallel to the line 4x − 2y − 5 = 0 if the slope of the tangent is equal to the slope of the line.

$\frac{3}{2 \sqrt{3 x-2}}=2$

$\Rightarrow \sqrt{3 x-2}=\frac{3}{4}$

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

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

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

When $x=\frac{41}{48}, y=\sqrt{3\left(\frac{41}{48}\right)-2}=\sqrt{\frac{41}{16}-2}=\sqrt{\frac{41-32}{16}}=\sqrt{\frac{9}{16}}=\frac{3}{4}$.

$\therefore$ Equation of the tangent passing through the point $\left(\frac{41}{48}, \frac{3}{4}\right)$ is given by,

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

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

$\Rightarrow 4 y-3=\frac{48 x-41}{6}$

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

$\Rightarrow 48 x-24 y=23$

Hence, the equation of the required tangent is.