Determine all the points of intrsection of the lines y=12/x and y^2=(3x^2+5)/2?

To determine the intercepting point of the curves (not
lines), we'll have to solve the system of equations of the
curves.

We'll write the
system:

2y^2 =3x^2+5

3x^2 -
2y^2 = -5 (1)

xy = 12 (2)

We
recognize an a homogenous system and the solving strategy is to eliminate the numbers
alone.

For this reason, we'll multiply the 1st equation by
12 and the 2nd equation by 5:

36x^2 - 24y^2 = -60
(3)

5xy = 60 (4)

We'll add (3)
and (4):

36x^2 + 5xy - 24y^2 =
0

We'll divide by x^2:

36 +
5y/x - 24y^2/x^2 = 0

We'll substitute y/x =
t:

-24t^2 + 5t + 36 = 0

24t^2
- 5t - 36 = 0

We'll apply quadratic
formula:

t1 =
[5+sqrt(25+3456)]/48

t1 =
(5+59)/48

t1 = 4/3

t2 =
(5-59)/48

t2 = -9/8

We'll put
y/x = 4/3

y = 4x/3

We'll
substitute y in the 2nd equation:

x*(4x/3) =
12

4x^2 = 4*9

x^2 =
9

x1 = 3 and x2 = -3

y1 = 4
and y2 = -4

y/x = -9/8

y =
-9x/8

-9x^2 = 8*12

-3x^2 =
32

x^2 = -32/3

This equation
has no real solutions.

The intercepting
points of the curves are represented by the following real solutions of the system, that
are: {(3 ; 4) ; (-3 ; -4)}.