Tangent and Normal Question3

In this page tangent and normal question3 we are going to see solution of some practice questions from the worksheet.

(2) Find the points on curve x² - y² = 2 at which the slope of the tangent is 2.

Solution:

slope of the tangent = 2

                         m = 2

x² - y² = 2

2 x - 2 y (dy/dx) = 0

     - 2 y (dy/dx) = -2 x

              dy/dx = x/y

slope of the tangent to the curve = x/y

                        x/y = 2

                           x = 2 y

to find the common point on the curve and tangent we have to apply x=2y in the given equation

        x² - y² = 2

      (2y)² - y² = 2

        4 y² - y² = 2

          3 y² = 2

             y² = 2/3

             y = ±√(2/3)

 y = √(2/3)              y = - √(2/3)

 x = 2√(2/3)             x = -2√(2/3)

Therefore the required points are [2√(2/3),√(2/3)] , [-2√(2/3),- √(2/3)]


(3) Find at what points on a circle x² + y² = 13, the tangent is parallel to the line 2x + 3 y = 7.

Solution:

Since the given line 2x + 3 y = 7 is parallel to the tangent line,slopes of two lines will be equal 

         m = - coefficient of x/coefficient of y

         m = -2/3

therefore the slope of tangent = -2/3  --- (1)

we can find slope of the tangent by differentiating the given equation of the curve.

x² + y² = 13

2 x + 2 y (dy/dx) = 0

      2 y (dy/dx) = -2 x

           dy/dx = -2 x/2y

           dy/dx = -x/y  --- (2)

              (1) = (2)

             -2/3 = -x/y

                2 y = 3 x

                  y = (3x/2)

Substitute y = (3x/2) in the equation of curve

 x² + y² = 13

 x² + (3x/2)² = 13

 x² + (9x²/4) = 13

  (4 x² + 9x²)/4 = 13

          13 x²/4 = 13

                x² = 13 (4/13)

                x² = 4

                x = ± 2

if x = 2                     if x = -2

y = 3(2)/2                   y = 3(-2)/2

y = 3                           y = -3  

therefore the required points are (2,3) (-2,-3)