Question

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.

Answer 1

HELLO DEAR,

The equation of the given curve is y = x² - 2x + 7.

differentiating w.r.t x, we get:
dy/dx = 2x - 2

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

2x − y + 9 = 0

⇒ y = 2x + 9

This is in the form of y = mx + c.

∴Slope of the line = 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

so, y = (2)² - 2(2) + 7
y = 4 - 4 + 7
y = 7

Thus, the equation of the tangent passing through (2, 7) is ;
y - 7 = 2(x - 2)

y - 2x = 3



Hence, the equation of the tangent line to the given curve (which is parallel to line 2x − y + 9 = 0) is $\boxed{\bold{y - 2x - 3 = 0}}$.


b) The equation of the line is 5y − 15x = 13.

5y − 15x = 13
⇒y = 3x + 15

This is in the form of y = mx + c.

∴Slope of the line = 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


2x - 2 = -1/3

2x = 5/3

x = 5/6

so, y = (5/6)² - 2(5/6) + 7

y = 25/36 - 10/6 + 7
y = (25 - 60 + 252)/36
y = 217/36

hence, the equation of the tangent passing through(5/6 , 217/36) is;

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

(36y - 217)/36 = -1/3×(6x - 5)/6

(36y - 217) = -2(6x - 5)

36y + 12x = 227

Hence,
the equation of the tangent line to the given curve (which is perpendicular to line 5y− 15x = 13) is $\boxed{\bold{36y + 12x = 227}}$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

HELLO DEAR,

The equation of the given curve is y = x² - 2x + 7.

differentiating w.r.t x, we get:

dy/dx = 2x - 2

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

2x − y + 9 = 0

⇒ y = 2x + 9

This is in the form of y = mx + c.

∴Slope of the line = 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

so, y = (2)² - 2(2) + 7

y = 4 - 4 + 7

y = 7

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

y - 7 = 2(x - 2)

y - 2x = 3

Hence, the equation of the tangent line to the given curve (which is parallel to line 2x − y + 9 = 0) is .

b) The equation of the line is 5y − 15x = 13.

5y − 15x = 13

⇒y = 3x + 15

This is in the form of y = mx + c.

∴Slope of the line = 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

2x - 2 = -1/3

2x = 5/3

x = 5/6

so, y = (5/6)² - 2(5/6) + 7

y = 25/36 - 10/6 + 7

y = (25 - 60 + 252)/36

y = 217/36

hence, the equation of the tangent passing through(5/6 , 217/36) is;

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

(36y - 217)/36 = -1/3×(6x - 5)/6

(36y - 217) = -2(6x - 5)

36y + 12x = 227

Hence,

the equation of the tangent line to the given curve (which is perpendicular to line 5y− 15x = 13) is

Step-by-step explanation: