tangent and normal question3

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)

SBI!