Tangent and Normal Question7





In this page tangent and normal question7 we are going to see solution of some practice questions from the worksheet.

(7) 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:

The tangent drawn at the point "P" on the curve touches the curve again at the point "Q"

Let P (a,a³) be a point on the curve

y = x³  --- (1)

dy/dx = 3 x²

slope at the point "P" (dy/dx) = 3 (a)²

                                        = 3 a²

Equation of the tangent at the point "P"

(y - y₁) = m (x - x₁)

 (y - a³) = 3 a² (x - a)

 y - a³ = 3 a² x - 3

now we are going to apply the value of y in this equation

x³ - a³ = 3 a² x - 3

x³ - a³ - 3 a² x + 3 a³ = 0

x³ - 3 a² x + 2 a³ = 0

(x - a)² (x + 2a) = 0

(x - a)² = 0                       (x + 2a) = 0

(x - a) = 0                                 x = - 2 a

     x = a

Here a is the x-coordinate value of the point "P". So we have to take -2a is the x-coordinate value of the point "Q"

Slope at the point Q

dy/dx = 3 x²

         = 3 (-2a)²

         = 3 (4a²)

         = 4 (3a²)

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


(8) 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

2 x₁² + 4 y₁² = 1     ----- (1)

6 x₁² - 12 y₁² = 1     ----- (2)

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

 (1) x 3 =>    6 x₁² + 12 y₁² = 3

                  6 x₁² - 12 y₁² = 1

                  ------------------

                   12 x₁² = 4

                      x₁² = 4/12

                      x₁² = 1/3

 now we are going to apply the value of x₁² in the first equation

2 (1/3) + 4 y₁² = 1

(2/3) + 4 y₁² = 1

          4 y₁² = 1 - 2/3

          4 y₁² = 1/3      

            y₁² = 1/12         

Therefore the common point on both curves is (1/3 , 1/12)

differentiate the first equation with respect to x

2 x² + 4 y² = 1

4 x + 8 y(dy/dx) = 0

8 y(dy/dx) = - 4 x

     (dy/dx) = - 4 x/8y

                = - x/2y

6 x₁² - 12 y₁² = 1

 12 x - 24 y (dy/dx) = 0

       - 24 y (dy/dx) = -12 x

                 dy/dx = -12 x/-24y

                         = x/2y

  If two curves are intersecting orthogonally then m₁ x m₂ = -1

(- x/2y) (x/2y) = -1

  - x²/4y² = -1

   -(1/3)/4(1/12) = -1   

 - (1/3)/(1/3) = - 1

          - 1 = -1

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