Find the equation of the tangent to the curve y = √3x - 2 which is parallel to the line 4x - 2y + 5 = 0

Solution:

The equation of the given curve is y = √3x - 2

The slope of the tangent to the given curve at any point (x, y) is given by,

dy/dx = 3/2 (√3x - 2)

The equation of the given line is 4x - 2y + 5 = 0

⇒ y = 2x + 5/2,

which is the form of y = mx + c

Hence, the slope of line is 2.

Now, the tangent to the given curve is parallel to the line 4x - 2 y + 5 = 0 if the slope of the tangent is equal to the slope of the line.

Therefore

3/2 √3x - 2 = 2

√3x - 2 = 3/4

⇒ 3x - 2 = 9/16

⇒ 3x = 9/16 + 2 = 41/16

⇒ x = 41/48

When, x = 41/48

Then,

y = √3 (41 / 48) - 2

= √41 / 16 - 2

= √(41 - 32) / 16

= √9 / 16

= 3/4

Thus, equation of the tangent passing through the point (41/48, 3/4) is

y - 3/4 = 2 (x - 41/48)

⇒ (4y - 3)/4 = 2 [(48x - 41) / 48]

⇒ (4y - 3) = (48x - 41)/6

⇒ 24y - 18 = 48x - 41

⇒ 48x - 24y = 23

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 25

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

Summary:

The equation of the tangent to the curve y = √3x - 2 which is parallel to the line 4x - 2y + 5 = 0 is 48x - 24y = 23