Question

Let p be the point on the curve y=x^3 and suppose that the tangent line at p intersect the curve again on q . prove that the slope of q is 4 times the slope of p

Answer 1

Answer:

Step-by-step explanation:

Let (a,a  

3

) be a point on the curve y=x  

3

 .....(1)

dx

dy

​  

=3x  

2

 

dx

dy

​  

 at (a,a  

3

)=3a  

2

=m= Slope of tangent

Equation of tangent at P is

y−a  

3

=3a  

2

(x−a)

(i.e.,) y−a  

3

=3a  

2

x−3a  

3

 

y=3a  

2

x−2a  

3

 .....(2)

Solving (1) and (2), we get

x  

3

=3a  

2

x−2a  

3

⇒x  

3

−3a  

2

x−2a  

3

=0

(x−a)  

2

(x+2a)=0

x=a or 2a

At x=a;y=a  

3

 at x=−2a,y=−8a  

3

 

But (a,a  

3

) is the point P

∴ Q=(−2a,−8a  

3

)

dx

dy

​  

 at Q (−2a,−8a  

3

)=3(−2a  

2

)=12a  

2

 

dx

dy

​  

 at P(a,a  

3

)=3(a  

2

)=3a  

2

 

∴ the slope of Q=4 (slope at P).