FIND ALL POINTS ON THE CURVE WHERE THE SLOPE OF TANGENT LINE IS GIVEN

Example 1 :

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  ------(1)

x²-y²  =  2

2x-2y(dy/dx)  =  0

-2y(dy/dx)  =  -2x

dy/dx  =  x/y

Slope of the tangent drawn at the point on the curve 

=  x/y  ------(2)

(1)  =  (2)

x/y = 2

x  =  2y

By applying the value of x in the given curve, we get

If y = √(2/3), then x  =  2√(2/3)

If y = -√(2/3), then x  =  -2√(2/3)

So, the required points are

(2√(2/3), √(2/3)) and (-2√(2/3), - √(2/3))

Example 2 :

Find at what points on a circle

x²+y²  =  13

the slope of the tangent is -2/3.

Solution :

m  =  -2/3  ----(1)

x² + y²  =  13

2x+2y(dy/dx)  =  0

2y(dy/dx)  =  -2x

dy/dx  =  -x/y  --- (2)

(1)  =  (2)

-2/3 = -x/y

2y  =  3x

y  =  3x/2

By applying y = 3x/2 in the given equation, we get

x² + (3x/2)²  =  13

 x² + (9x²/4)  =  13

13x²/4  =  13

x²  =  13(4/13)

x  =  ± 2

If x = 2, then y = 3 

If x = -2, then y  =  -3

So, the required points are (2, 3) (-2, -3).

Example 3 :

For the curve y = 4x³ -2x⁵, find all the points at which the tangent passes through the origin.

Solution :

y = 4x³ -2x⁵

Slope (dy/dx) = 4(3x²) - 2(5x⁴)

= 12x² - 10x⁴

since tangent is taken from (h, k). Slope of tangent at (h, k) is 

dy/dx = 12h² - 10h⁴

Equation of tangent :

(y - y₁) = m(x - x₁)

y - k = (12h² - 10h⁴)(x - h)

It passes through origin, then

y - k = (12h² - 10h⁴)(x - h)

0 - k = (12h² - 10h⁴)(0 - h)

k = (12h² - 10h⁴)h

k = 12h³ - 10h⁵ -----(1)

The curve passes through the point (h, k).

k = 4h³ -2h⁵ -----(2)

(1) - (2)

12h³ - 10h⁵ = 4h³ -2h⁵

12h³ - 4h³- 10h⁵ + 2h⁵ = 0

8h³- 8h⁵ = 0

8h³(1 - h²) = 0

h = 0 and 1 - h² = 0

h = 1 and -1

So, the values of h are 0, 1, -1.

k = 4h³ -2h⁵

So, the required points are (0, 0) (-1, -2) and (1, 2).

Example 4 :

The curve

y = ax³ + bx² + cx + 5

touches the x-axis at the point (-2, 0) and Cuts the y-axis at a point where the slope is 3. Find a, b, c

Solution :

y = ax³ + bx² + cx + 5

dy/dx = a(3x²) + b(2x) + c(1) + 0

dy/dx = 3ax² + 2bx + c

Slope is 3, then 3ax² + 2bx + c = 3

Cuts the y-axis, when slope = 3

3a(0)² + 2b(0) + c = 3

c = 3

y = ax³ + bx² + cx + 5

The curve touches at (-2, 0)

0 = a(-2)³ + b(-2)² + c(-2) + 5

-8a + 4b - 2c = -5

-8a + 4b - 2(3) = -5

-8a + 4b = -5 + 6

-8a + 4b = 1 ------(1)

The tangent touches the x-axis at (-2, 0).

dy/dx = 3ax² + 2bx + c

0 = 3a(-2)² + 2b(-2) + c

0 = 12a - 4b + 3

12a - 4b = -3 ----(2)

(1) + (2)

-8a + 4b + 12a - 4b = 1 + (-3)

4a = -2

a = -1/2

Applying the value of a in (1), we get

-8(-1/2) + 4b = 1

4 + 4b = 1

4b = 1 - 4

4b = -3

b = -3/4

So, the values of a = -1/2, b = -3/4 and c = 3.

Example 5 :

The point on the curve y = x³ – 11x + 5 at which the tangent is y = x –11 is

Solution :

y = x³ – 11x + 5

Equation of tangent is y = x - 11

Comparing with y = mx + b, we get slope = 1

By finding derivative,

dy/dx = 3x² - 11(1) + 0

3x² - 11 = 1

3x² = 12

x² = 12/3

x² = 4

x = 2 and -2

When x = -2, y = (-2)³ – 11(-2) + 5

= -8 + 22 + 5

= -8 + 27

= 19

When x = 2, y = 2³ – 11(2) + 5

= 8 - 22 + 5

= 13 - 22

= 9

So, the required points are (-2, 19) and (2, 9).

Example 6 :

The line y = x + 1 is a tangent to the curve y² = 4x at the point

(a) (1, 2)     (b) (2, 1)    (c) (1, – 2)   (d) (–1 ,2)

Solution :

y = x + 1

slope = 1

y² = 4x

2y (dy/dx) = 4(1)

dy/dx = 4/2y

4/2y = 1

2/y = 1

y = 2

applying y = 2 in y² = 4x, we get

4 = 4x

x = 1

So, the required point is (1, 2). So, option a is correct.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.