PROBLEMS ON TANGENTS AND NORMALS

Problem 1 :

Let P be a point on the curve

y  =  x³

and suppose that the tangent line at P intersects the curve again at Q. Prove that the slope at Q is four times the slope at P.

Solution :

Let P (a, a³) be a point on the curve.

y  =  x³  --- (1)

dy/dx = 3x²

Slope at the point P :

dy/dx  =  3a²

Equation of the tangent at the point P : 

(y-y₁)  =  m(x-x₁)

(y-a³)  =  3a²(x-a)

 y-a³  =  3a²x-3a³

By applying the value of y  =  x³

x³-a³  =  3a²x-3a³

x³-a³-3a²x+3a³

x³-3a²x+2a³  =  0

(x-a)² (x+2a)  =  0

Slope at the point Q :

dy/dx  =  3x²

= 3(-2a)²

= 3(4a²)

= 4(3a²)

Slope at the point Q  =  4(Slope at the point P)

Problem 2 :

Prove that the curve

2x²+4y²  =  1 and 6x²-12y²  =  1

cut each other at right angles. 

Solution :

Let (x₁, y₁) be the common point on the curve

2x₁²+4 y₁²  =  1     ----- (1)

6x₁²-12y₁²  =  1     ----- (2)

to find the point of intersection we have to solve the given equations

 (1) x 3 =>    6x₁²+12y₁²  =  3

                  6x₁²-12y₁²  =  1

                  --------------------

                   12x₁²  =  4

x₁²  =  4/12 ==>  x₁²  =  1/3

By applying the value of x₁² in (1), we get

2(1/3)+4y₁²  =  1

(2/3)+4y₁²  =  1

4y₁²  =  1 - 2/3

4y₁²  =  1/3

y₁²  =  1/12

Point of intersection is (1/3 , 1/12).

2x²+4y²  =  1

4x+8y(dy/dx)  =  0

8y(dy/dx)  =  -4x

dy/dx  =  -4x/8y

dy/dx   = -x/2y

6x₁²-12y₁²  =  1

12x-24y(dy/dx)  =  0

dy/dx  = x/2y

If two curves are intersecting orthogonally then

m₁ ⋅ m₂  =  -1

(- x/2y) (x/2y)  =  -1

-x²/4y²  =  -1

-(1/3)/(1/3)  =  -1   

-(1/3)/(1/3)  = -1

- 1  =  -1

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