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)