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

Explanation

Equation of curve
`y = sqrt(3x-2)` …(1)
` rArr (dy)/(dx) = 1/(2sqrt(3x-2)) d/(dx) (3x-2)`
` rArr (dy)/(dx) = 3/(2sqrt(3x-2))`
Slope of tangent at point `(x, y)= 3/(2 sqrt(3x-2))`
equation of line ` 4x- 2y + 5 = 0`
` rArr y= 2x + 5/2`
slope of line = 2
Given that ` 3/(2sqrt(3x-2)) = 2`
` rArr sqrt(3x-2) = 3/4 rArr 3x - 2 = 9/16`
` rArr 3x = 41/16 rArr x = 41/48`
put x = ` 41/48` in equation (1)
`y = sqrt(3 xx (41)/(48) - 2) = 3/4`
Equation of tangent at point `(41/48 , 3/4)`
` y-3/4 = 2 (x - 41/48)`
` rArr (4y-3)/4 = (48x-41)/24`
` rArr 24y - 18 = 48 x - 41 `
` rArr 48 x - 24y - 23 = 0`