Question

Find the equation of tangent to the curve `y = sqrt(3x -2)` which is parallel to the line 4x − 2y + 5 = 0. Also, write the equation of normal to the curve at the point of contact.

Answer 1

Slope of the given line is 2
Let (x₁, y₁) be the point where the tangent is drawn to the curve `y = sqrt(3x -2)`
Since, the point lies on the curve.
Hence, `y_1 = sqrt(3x_1-2)`   ...(1)
Now, `y = sqrt(3x -2)`
⇒ `dy/dx = 3/(2sqrt(3x-2)`
Slope of tangent at `(x_1,y_1) = 3/(2sqrt(3x_1-2)`

Given that
Slope of tangent = slope of the given line
⇒ `3/(2sqrt(3x_1-2)` = 2
⇒  `3 = 4 sqrt(3x_1-2)`
⇒  `9 = 16 (3x_1 - 2)`
⇒  `9/16 = 3x_1 -2`

⇒  `3x_1 = 9/16 + 2 = (9+32)/16 = 41/16`

⇒  `x_1 = 41/48`

Now,
`y_1 = sqrt(123/48 -2) = sqrt(27/48) = sqrt(9/16) = 3/4`  ...[From (1)]

∴ `(x_1,y_1) = (41/48,3/4)`
Equation of tangent is,
`y - y_1 = m (x -x_1)`

⇒  `y - 3/4 = 2 (x-41/48)`

⇒  `(4y-3)/4 = 2 ((48x -41)/48)`

⇒  24y - 18 = 48x - 41

⇒  48x - 24y - 23 = 0

Equation of normal at the point of contact will be
`y-y_1 = -1/m (x -x_1)`

⇒ `y - 3/4 = -1/2 (x -41/48)`

⇒ `(4y-3)/4 = -1/2 (x -41/48)`

⇒ `(4y-3)/2 = (41/48 -x)`

⇒ `(4y-3)/2 = (41-48x)/48`

⇒ `4y - 3 = (41 -48x)/24`

⇒ 96y - 72 = 41 -  48x

⇒ 48x + 96y = 113