# 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.

Given$y = \sqrt{3x-2}$

Differentiating both sides w.r.t x we get

$\frac{dy}{dx} = \frac{3}{2\sqrt{3x-2}} = m_1$

and equation of the line $4x -2y + 5 = 0$

$Slope \ m_2 = \frac{- \textup{coeff of } x }{\textup{coeff of }y} = \frac{-4}{-2} = 2$

Applying condition of parallel line

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

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

$9= 16(3x-2) \Rightarrow 9 = 48x -32$

$41 = 48x$

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

Substituting value of $x$ in (1)

$y = \sqrt{3\frac{41}{48}-2} \Rightarrow \frac{3}{4}$

$\therefore$ Point of contact $\left(\frac{41}{48},\frac{3}{4} \right )$

Slope of curve = 2

So Equation of tangent is

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

$\left ( \frac{4y -3}{4} \right ) = {\left ( \frac{48x - 41}{24} \right )}$

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

${24y -18} = {48x - 41}$

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

Equation of normal is:

${ \left (y - \frac{3}{4} \right )} = -\frac{1}{2}{\left (x - \frac{41}{48} \right )}$

$24(4y-3) = -48x + 41$

$96y - 72 = -48x +41$

$96 y +48x = 41 + 72$

$48x + 96y -113 =0$

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