Find the equation of tangent to the curve y = \sqrt{3x -2} which is parallel to the line4x -2y +5 = 0. Also, write the equation of normal to the curve at the point of contact.

 

 

 

 
 
 
 
 

Answers (1)

Giveny = \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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