Application of Derivatives — Class 12 Maths Solution

We have,
$y = \sqrt {3x - 2}$ …(i)
and $4x - 2y + 5 = 0$ …(ii)
Slope of the line (ii) is $2$

From (i),$\cfrac{{dy}}{{dx}} = \cfrac{3}{{2\sqrt {3x - 2} }}$

Let $({x_1},{y_1})$ be the point on (i) at which tangent is parallel to
(ii), then
${\left( {\cfrac{{dy}}{{dx}}} \right)_{({x_1},{y_1})}} =$Slope of line (ii)

$\Rightarrow \cfrac{3}{{2\sqrt {3{x_1} - 2} }} = 2 \Rightarrow 3 = 4\sqrt {3{x_1} - 2}$

$\Rightarrow 3{x_1} - 2 = {\left( {\cfrac{3}{4}} \right)^2} \Rightarrow {x_1} = \cfrac{{41}}{{48}}$

Also, $({x_1},{y_1})$ lies on (i),

Therefore,
${y_1} = \sqrt {3{x_1} - 2} = \sqrt {3x\cfrac{{41}}{{48}} - 2} . = \sqrt {\cfrac{{123 - 96}}{{48}}} = \sqrt {\cfrac{{27}}{{48}}} = \sqrt {\cfrac{9}{{16}}} = \cfrac{3}{4}$

Therefore the point on (i) at which tangent is parallel to (ii) is $\left( {\cfrac{{41}}{{48}},\cfrac{3}{4}} \right)$.

Therefore the required equation of tangent is
$y - \cfrac{3}{4} = 2\left( {x - \cfrac{{41}}{{48}}} \right)$ or $48x - 24y - 23 = 0$