#### To determine

**To show:** If two lines L1 and L2 intersect at an angle Α, then

tanΑ = m2 - m11 + m1m2 where m1 and m2 are the slopes of L1 and L2 respectively, by using identity for tan(x-y)

#### Answer

tanΑ = m2 - m11 + m1m2

#### Explanation

**1) Concept:**

tan(x-y) = tanx - tany1 +tanxtany

If two lines L1 and L2 have slopes m1 and m2 and angles of inclination θ1 and θ2 shown in the following figure.

Here, slope of line L1 is m1 = tanθ1 and slope of line L2 is m2 = tanθ2

**2) Given: **

Two lines L1 and L2 intersect at an angle Α, then we need to show that

tanΑ = m2 - m11 + m1m2

where m1 and m2 are the slopes of L1 and L2, respectively

3) **Calculation:**

In triangle ABC, A+ B+ C =1800

Α+ θ1+ 1800- θ2 =1800

Subtract 1800 from both sides,

Α+ θ1 - θ2 = 0

By adding θ2 on both sides,

Α+ θ1 = θ2

Then, subtract θ1 from both sides,

Α = θ2- θ1

Now, take tan of both sides,

tan Α= tan(θ2- θ1)

By using identity for tan(θ2- θ1),

tan Α= tan(θ2- θ1)= tanθ2 - tanθ11 +tanθ2tanθ1

Substitute tanθ2=m2 and tanθ1=m1 in above step,

tan Α= tan(θ2- θ1)= m2 - m11 + m2m1

Therefore,

tan Α= m2 - m11 + m2m1

Hence verified

**Conclusion:**

Therefore,

tan Α= m2 - m11 + m2m1

#### To determine

**Part (b):**

**To find: **The angle between each pair of curves at each point of intersection correct to the nearest degree using part (a).

#### Answer

The angle between the curves C1 and C2 at each point of intersection:

i) Α 530 or Α11270

ii) Α 630 or Α11170

#### Explanation

**1) Given:**

i) C1 : y=x2 and C2 : y=(x-2)2

ii) C1 : y : x2-y2=3 and C2 : y= x2-4x+y2+3=0

**2) Calculations:**

**i) **To find point of intersections of C1 and C2 equate right side of both equations and solve for x

x2 = (x-2)2

By using FOIL method on right side,

x2 = x2-4x+4

Subtract x2 on both sides,

0 = -4x+4

Now, add 4x on both sides

4x= 4

Divide both sides by 4,

x=1

To find y value, substitute x=1 in any equation and solve for y.

y=1

Thus, the intersection point Pis (1, 1).

Calculate slope of each given curve C1 and C2 at point P(1, 1).

Differentiate curve C1 and curve C2 with respect to x,

y= x2

dydx=2x

To find the slope of the tangent m1 at (1, 1), substitute x = 1 in above derivative.

So, m1=2

Similarly, differentiate y=(x-2)2 with respect to x

dydx=2(x-2)

To find the slope of the tangent m2 at (1, 1), substitute x = 1 in above derivative.

So, m2=-2

So the slope of the tangents to C1 and C2 at (1, 1) are m1= 2 and m2= -2 respectively.

From part (a),

tan Α= m2 - m11 + m2m1

Substitute, m1= 2 and m2= -2 in above step

tanΑ= -2 -21 +-2*2 = 43

tanΑ=43

By taking tan-1 of both sides,

Α= tan-143530 ………………..(1)

To find second angle between given curves, subtract 530 from 1800

Α11800-530 1270 …………………(2)

**ii)**C1 : x2-y2=3 and C2 : x2-4x+y2+3=0

From first equation, find y2 which is y2= x2-3

To find point of intersections, substitute y2= x2-3 in x2-4x+y2+3=0

x2-4x+x2-3+3=0

2x2-4x=0

Factor out 2x from left side of the above step,

2x(x-2)=0

x=0 or x = 2

Here, x=0 is extraneous since second equation is equation of circle with radius 1and center (2, 0).

Now find y value by substituting x = 2 in either equation of part (ii).

y2=x2- 3

y2=1

By using square root property,

y= 1

Thus, the intersection points (2, 1) and (2, -1)

Calculate slope of each given curve C1 and C2 at both points(2, 1) and (2, -1).

Differentiate curve C1 and curve C2 with respect to x,

x2-y2=3

2x-2y dydx =0

Divide both sides by 2,

x-y dydx =0

Subtract x from both sides,

-y dydx =-x

Divide both sides by -y,

dydx = xy

To find the slope of the tangent m1 at (2, 1), substitute x = 2 and y = 1 in above derivative.

So, m1=2

Similarly, differentiate x2-4x+y2+3=0 with respect to x

2x-4+2ydydx =0

Divide both sides by 2, since 2 is common in all terms on the left side of the above equation

x-2+ydydx =0

Subtract x from both sides,

-2+ydydx =-x

Now, add 2 on both sides

ydydx =2-x

Divide both sides by y,

dydx =2-xy

To find the slope of the tangent m2 at (2, 1), substitute x = 2 in above derivative.

So, m2=0

The slope of the tangents to C1 and C2 at (2, 1), m1= 2 and m2= 0 respectively

From part (a),

tan Α= m2 - m11 + m2m1

Substitute, m1= 2 and m2= 0 in above step

tanΑ= 0 -21 +0*2 = -2

tanΑ=-2

By taking tan-1 of both sides,

Α= tan-1-21170 …………….(3)

To find second angle between given curves, subtract 1170 from 1800

Α11800-1170 630 ……………..(4)

To find the slope of the tangent m1 at (2, -1), substitute x = 2 and y = -1 in derivative of C1

dydx = xy

So, m1=-2

Similarly, find slope of the tangent m2 at (2, -1), substitute x = 2 and y = -1 in derivative of C2

dydx =2-xy

So, m2=0

From part (a),

tan Α= m2 - m11 + m2m1

Substitute, m1= -2 and m2= 0 in above step

tanΑ= 0 -(-2)1 +0*(-2) = -2

tanΑ=2

By taking tan-1 of both sides,

Α= tan-12630 …………………(5)

To find second angle between given curves, subtract 630 from 1800

Α11800-630 1170 …………………(6)

From (1), (2), (3), (4), (5) and (6)

The angle between the curves C1 and C2 at each point of intersection:

i) Α 530 or Α11270

ii) Α 630 or Α11170

**Conclusion:**

Therefore, the angle between the curves C1 and C2 at each point of intersection:

i) Α 530 or Α11270

ii) Α 630 or Α11170