Question

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

Answer 1

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

The slope of the tangent at any point (x,y) is given by,

$\frac{dy}{dx}=\frac{1}{2}(3x-2{)}^{\frac{1}{2}-1}.3=\frac{3}{2\sqrt{3x-2}}$

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

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

$\therefore$ Slope of this line is 2.

Now, the tangent to the curve is parallel to the line 4x-2y+5=0 i.e., the slope of the tangent is equal to the slope of the line.

$\Rightarrow \frac{3}{2\sqrt{3x-2}}=2\Rightarrow 4\sqrt{3x-2}\Rightarrow 3x-2={\left(\frac{3}{4}\right)}^2\Rightarrow 3x=2+\frac{9}{16}\Rightarrow x=\frac{41}{48}$

Now, putting $x=\frac{41}{48}$ in $y=\sqrt{3x-2}$ , we get

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

$\therefore$ Equation of the tangent passing throught the point $(\frac{41}{48},\frac{3}{4})$ having slope 2 is given by

$y-\frac{3}{4}=2(x-\frac{41}{48})\Rightarrow y-\frac{3}{4}=2x-\frac{41}{24}\Rightarrow 48x-24y=23$

Hence, the equation of the required tangent is 48x-24y=23.