# Find the Equation of Tangent to the Curve Y = √ 3 X − 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. - Mathematics

Sum

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.

#### Solution

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

Concept: Tangents and Normals
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