Question

the slope of the tangent of the curve y^2 = x^3 at the point (4,8) is
​

Answer 1

Answer:

104

Step-by-step explanation:

Solving the curve at x=8=2

3

y

2

=x

3

=2

9

⇒y=±2

4

2

=±16

2

So the points are (8,±16

2

)

Now the curve is, y

2

=x

3

⇒2y×

dx

dy

=3x

2

⇒(

dx

dy

)=

2y

3x

2

⇒(

dx

dy

)

(8,±16

2

)

=

±2×16

2

3×64

=±3

2

Thus the slope of normal is =(

dx

dy

)=∓

3

2

1

Therefore, equation of normal is,

(y∓16

2

)=∓

3

2

1

(x−8)

⇒∓3

2

y±96=(x−8)

⇒x±3

2

y=104