(a) It is given that equation of the curve is y = x² – 2x +7

On differentiating with respect to x, we get

The equation of the line is 2x – y + 9 = 0

⇒ y = 2x + 9

⇒ Slope of the line = 2

Now we know that if a tangent is parallel to the line 2x – y + 9 = 0, then

Slope of the tangent = Slope of the line

⇒ 2 = 2x – 2

⇒ 2x = 4

⇒ x = 2

Now, putting x = 2, we get

y =4 -4 + 7 = 7

Then, the equation of the tangent passing through (2,7)

⇒ y – 7 = 2(x – 2)

⇒ y – 2x – 3 = 0

Therefore, the equation of the tangent line to the given curve which is parallel to line 2x – y + 9 = 0 is y – 2x – 3 = 0.

(b) It is given that equation of the curve is y = x² – 2x +7

On differentiating with respect to x, we get

The equation of the line is 5y – 15x = 13

⇒ y =

⇒ Slope of the line = 3

Now we know that if a tangent is perpendicular to the line 5y – 15x = 13, then

⇒ 2x – 2=

⇒ 2x =

⇒ x =

Now, putting x = , we get

y =

Then, the equation of the tangent passing through

⇒ y – = (x – )

⇒

⇒ 36y – 217 = -2(6x -5)

⇒ 36y+12x – 227 = 0

Therefore, the equation of the tangent line to the given curve which is perpendicular to line 5y – 15x = 13 is 36y+12x – 227 = 0.