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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.

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

Answers (1)

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}

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