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#### Provide Solution For R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 16 Maths Textbook Solution.

Answer: Equation of the tangent is $\Rightarrow 48x-24y=23$

Hint:

Given: $y=\sqrt{3x-2}$ Which is paralle to $yx-2y+5=0$

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

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

$\frac{dy}{dx}=\frac{2}{2\sqrt{3x-2}}$ The Equation of the given line is $4x-2y+5=0$

$4x-2y+5=0,\therefore y=2x+\frac{5}{2}$     $\left [ which \; is \; in \; the \: form \: of\; y =mx+c \right ]$

$\therefore slope\: of \: line \: is\: 2$

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

\begin{aligned} &\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} \\ &\text { When } x=\frac{41}{48}, y=\sqrt{3\left[\frac{41}{48}\right]-2}=\frac{3}{4} \end{aligned}

the equation of tangent at point $\left [ \frac{41}{48},\frac{3}{4} \right ]$ is given by

\begin{aligned} &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 48 x-24 y=23 \end{aligned}