# Find the equation of the tangent line to the curve y = x^{2} - 2x + 7 which is

(a) parallel to the line 2x - y + 9 = 0

(b) perpendicular to the line 5y - 15x = 13

**Solution:**

The equation of the curve is y = x^{2} - 2x + 7

On differentiating with respect to x , we get:

dy/dx = 2x - 2

(a) The equation of the line is 2x - y + 9 = 0

⇒ y = 2x + 9

This is of the form y = mx + c

Hence, the slope of line is 2

If a tangent is parallel to the line 2x - y + 9 = 0,

then the slope of the tangent is equal to the slope of the line.

Therefore, we have:

2 = 2x - 2

⇒ 2x = 4

⇒ x = 2

Now, x = 2

Then,

⇒ y = 4 - 4 + 7

⇒ y = 7

Thus, the equation of tangent passing through (2, 7) is given by,

y - 7 = 2 (x - 2)

⇒ y - 2x - 3 = 0

(b) perpendicular to the line 5y - 15x = 13

⇒ y = 3x + 13/5

This is of the form y = mx + c

Hence, the slope of line is 3

If a tangent is perpendicular to the line 5y - 15x = 13,

then the slope of the tangent is,

- 1 / slope of the line

= (- 1)/3

Therefore, we have:

2x - 2 = - 1/3

⇒ 2x = - 1/3 + 2

⇒ 2x = 5/3

⇒ x = 5/6

Now,

x = 5/6

Then,

y = 25/36 + 10/6 + 7

= (25 - 60 + 252) / 36

= 217/36

Thus, the equation of tangent passing through (5/7, 217/36) is given by:

y - 217/36 = 1/3 (x - 5/6)

⇒ (36y - 217) / 36

= - 1/18 (6x - 5)

⇒ 36y - 217 - 2 (6x - 5)

⇒ 36y - 217 = - 12x + 10

⇒ 36y + 12x - 227 = 0

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 15

## Find the equation of the tangent line to the curve y = x^{2} - 2x + 7 which is (a) parallel to the line 2x - y + 9 = 0 (b) perpendicular to the line 5y - 15x = 13

**Summary:**

The equation of the tangent line to the curve y = x^{2} - 2x + 7 which is parallel to the line 2x - y + 9 = 0 is y - 2x - 3 = 0 and perpendicular to the line 5y - 15x = 13 is 36y + 12x - 227 = 0

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