Question

Find the equation of tangent to the curve $y=\sqrt{3x-2}$ which is parallel to the line $4x-2y+5=0.$ Also, write the equation of normal to the curve at the point of contact.

Answer 1

$y=\sqrt{3x-2}$
$\Rightarrow \frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}$
Also, the equation of the line parallel to the tangent is $4x-2y+5=0$
$\Rightarrow 2y=4x+5$
$\Rightarrow y=2x+\frac{5}{2}$
Therefore, the slope of the line is $2.$

As tangent to the curve and the given line are parallel, their slopes are equal.
$\Rightarrow \frac{3}{2\sqrt{3x-2}}=2$
$\Rightarrow \frac{3}{4}=\sqrt{3x-2}$
$\Rightarrow 3x-2=\frac{9}{16}$
$\Rightarrow x=\frac{41}{48}$

$\text{For}x=\frac{41}{48},$ the value of $y$ for the curve $y=\sqrt{3x-2}$ is
$y=\sqrt{3\times \frac{41}{48}-2}$
$\Rightarrow y=\sqrt{\frac{41}{16}-2}\Rightarrow y=\sqrt{\frac{9}{16}}\Rightarrow y=\frac{3}{4}$

Equation of the tangent through $(\frac{41}{48},\frac{3}{4})$ is
$y-\frac{3}{4}=2(x-\frac{41}{48})\Rightarrow 48y-36=2(48x-41)$
$\Rightarrow 48y-36=96x-82\Rightarrow 96x-48y-46=0$

Slope of the normal at the point $(\frac{41}{48},\frac{3}{4})$ is $\frac{-1}{\text{slope of tangent}}=\frac{-1}{2}$
Therefore, equation of normal is
$y-\frac{3}{4}=\frac{-1}{2}(x-\frac{41}{48})\Rightarrow 48y-36=\frac{-1}{2}(48x-41)$
$\Rightarrow 96y-72=-48x+41\Rightarrow 48x+96y-113=0$

Therefore, required equation of the tangent is $96x-48y-46=0$ and equation of the normal is $48x+96y-113=0$