Question

find the equations of tangent and normal to the curve y= 3x² - 3x -5 where the tangent is parallel to the line 3x - y + 1 = 0​

Answer 1

Answer:

Step-by-step explanation:

I attached the photos of answer may it help you mate

Answer 2

Step-by-step explanation:

SOLUTION

We have equation of the curve 3x2- y2

= 8

Differentiating both sides w.r.t. x, we get

6x - 2y

dy

= - 6X

dx

dy 3xt

dx

Slope of the tangent to the given curve

Slope of the normal to the curve =

9

3

3c

Now differentiating both sides the given

line x+3y = 4

1+3. =0

dx

dx

Since the normal to the curve is parallel

to the given line x + 3y =4

1

3

y X

Putting the value of y in 3x2- y2 8, we

get

3x2- x2=8

2x2 8

x= 4

X = t2

y=t2

The points on the curve are (2, 2) and

(-2,-2)

Now equation of the normal to the

curve at (2, 2) is

y-2 -2)

3y 6= -x+2

X+3y = 8

At (-2-2)y+2-+2)

3y +6 -x- 2

X+3y =- 8

Hence, the required equations are x +3y

=8 and x +3y = - 8 orx+3y = +8.