## 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)