Question

Find the measure of the angle between x²-y²=3, x²+y²-4x+3=0

Answer 1

angle between two curve : find slope of both curves . Let $m_1$ and $m_2$ are the slopes of given curves.
then, angle between them =$\tan^{-1}\frac{|m_2-m_1|}{|1+m_1.m_2|}$

curve, x² - y² = 3
differentiate with respect to x,
2x - 2y.dy/dx = 0
dy/dx = x/y ..........(i)

curve, x² + y² -4x + 3 = 0
differentiate with respect to x,
2x + 2y.dy/dx - 4 = 0
dy/dx = (4 - 2x)/2y = (2 - x)/y ......(ii)

now slove curves (1) and (2),
x² - y² = 3 ...........(1)
x² + y² -4x + 3 = 0.......(2)
adding equations (1) and (2),
2x² - 4x + 3 = 3
2x² - 4x = 0
x = 0, 2
put x = 0, in x² - y² = 3 => y² ≠ -3
so, x = 0, is not a solution .

again, put x = 2 in x² - y² = 3 => 4 - y² = 3
y² = 1 => y = ±1

so, intersecting points are (2, 1) and (2, -1)

at (2,1) equation (i),
slope of 1st curve , $m_1$ = 2

at (2,1) equation (ii),
slope of 2nd curve , $m_2$ = 0

then, angle between them, $\theta$
= $tan^{-1}\frac{2-0}{1+2.0}$
= $tan^{-1}2$

[note :- you can get same angle if you put (2,-1) ]

Answer 2

Dear Student:

Angle between two curve can be given by  finding  slope of both curves .  

1st curve, x² - y² = 3

differentiate with respect to x,

2x - 2y.dy/dx = 0

dy/dx = x/y ..........(i)

2nd curve, x² + y² -4x + 3 = 0

differentiate with respect to x,

2x + 2y.dy/dx - 4 = 0

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

Solve curves (1) and (2),

x² - y² = 3 ...........(1)

x² + y² -4x + 3 = 0.......(2)

(1) +(2),

2x² - 4x + 3 = 3

2x² - 4x = 0

x = 0, 2

put x = 0, in x² - y² = 3 => y² ≠ -3

So, x = 0, is not a solution .

Again, put x = 2 in x² - y² = 3 => 4 - y² = 3

y² = 1 => y = ±1

Hence, intersecting points are (2, 1) and (2, -1)

at (2,1) equation (i),

slope of 1st curve ,  = 2

slope of 2nd curve ,  = 0

Then, angle between them,

[note :-We will get same angle if we put (2,-1) ]

See the attachment: