Find the equation of tangent to the curve y = √ (3x-2) , which is parallel to the line 4x – 2y + 5 = 0. ​

Question

Find the equation of tangent to the curve y = \(\sqrt{3x-2}\) , which is parallel to the line 4x – 2y + 5 = 0.

Answer 1

Given,

y = \(\sqrt{3x-2}\)

⇒ y² = 3x - 2 ...(i)

To get the equation of tangent to the curve y2 = 3x - 2, which is a parabola, first we have to find the coordinates of point from where tangent line passes.

Let the coordinates of the point on parabola be (a,b),then this coordinate will satisfy the equation (i).

Therefore, from (i), we have 

b² = 3a - 2 ..(ii)

Now differentiating the equation (i) with respect to x, we get

\(2y\frac{dy}{dx}=3\) 

⇒ \((\frac{dy}{dx})=\frac{3}{2y}\)

Now slope of the required tangent line,

m₁ = \((\frac{dy}{dx})_{(a,b)}=\frac{3}{2b}\)

As it is given that required tangent line is parallel to the given line 4x - 2y + 5 = 0, the slopes of lines are equal.

Therefore,

\(\frac{3}{2b}=\frac{-4}{-2}\) 

⇒ 4b = 3

⇒ b = \(\frac{3}{4}\) 

Substituting this value b = \(\frac{3}{4}\) in equation (ii), we get

(\(\frac{3}{4}\))² +2 = 3a

⇒ 3a = \(\frac{41}{16}\)  

⇒ a = \(\frac{41}{48}\)  

Now the coordinates of the point on tangent are \(\frac{41}{48}\), \(\frac{3}{4}\) and slope is 2.

Hence, equation of tangent is obtained by,

y - b = m(x - a)

⇒ 48x - 24y - 23 = 0 is the required equation of tangent.