## Tangent and Normal Question4

In this page tangent and normal question4 we are going to see solution of some practice questions from the worksheet.

(4) At what points on the curve x²+y²-2x-4y+1 = 0 the tangent is parallel to

(i) x - axis   (ii) y - axis

Solution:

Since the tangent is parallel to x - axis, then slope of line will be 0.

m (or) (dy/dx) = 0

2 x + 2 y(dy/dx) - 2 (1) - 4 (dy/dx) + 0 = 0

2 x + 2 y(dy/dx) - 2 - 4 (dy/dx) + 0 = 0

2 y(dy/dx) - 4 (dy/dx) + 0 = - 2 x + 2

(dy/dx) [2 y - 4] = 2 (-x + 1)

(dy/dx) = [2 (-x + 1)]/[2(y - 2)]

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

(-x + 1)/(y - 2) = 0

- x + 1 = 0

- x = -1

x = 1

now we are going to apply the value x = 1 in the given equation

x²+y²-2x-4y+1 = 0

(1)² + y² - 2 (1) - 4 y + 1 = 0

1 - 2 + y² - 4 y + 1 = 0

2 - 2 + y² - 4 y = 0

y (y - 4) = 0

y = 0       y = 4

therefore the required points of contact are (1,0) and (1,4)

(ii) y - axis

Solution:

Since the tangent is parallel to y - axis, then slope of line will be 1/0.

m (or) (dy/dx) = 1/0

differentiate with respect to y

x² + y² - 2 x - 4 y + 1 = 0

2 x (dx/dy) + 2 y - 2 (dx/dy) - 4 (1) + 0 = 0

2 x (dx/dy) + 2 y - 2 (dx/dy) - 4 = 0

2 x (dx/dy) - 2 (dx/dy) = 4 - 2 y

(dx/dy) [2 x - 2] = 2 (2 - y)

dx/dy = [2 (2 - y)]/[2 (x - 1)]

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

0/1 = (2 - y)/(x - 1)

2 - y = 0

y = 2

Now apply y = 2

x² + (2)² - 2 x - 4 (2) + 1 = 0

x² + 4 - 2 x - 8 + 1 = 0

x² - 2 x - 3 = 0

(x - 3) (x + 1) = 0

x - 3 = 0     x + 1 = 0

x = 3            x = -1

Therefore the required points are (3,2) (-1,2)

tangent and normal question4 tangent and normal question4