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

justaguide | Certified Educator

At the points where the curves y = 12/x and y^2 = (3x^2 + 5)/2 intersect, the values of the x and y coordinates are the same.

y = 12/x

=> y^2 = 144/x^2

y^2 = (3x^2 + 5)/2

substitute y^2 = 144/x^2

144/x^2 = 3(x^2)/2 + 5/2

=> 288 = 3x^4 + 5x^2

=> 3x^4 + 5x^2 - 288 = 0

=> 3x^4 + 32x^2 - 27x^2 - 288 = 0

=> x^2(3x^2 + 32) - 9(3x^2 + 32) = 0

=> (x^2 - 9)(3x^2 + 32) = 0

x^2 = 9 => x = 3, x = -3

3x^2 + 32 = 0 => x = sqrt (-32/3) and sqrt (32/3)

For the points of intersection we can ignore the complex values.

y = 12/x => y = 4 for x = 3 and y = -4 for x = -3

The points of intersection are (4,3) and (-4,-3)

giorgiana1976 | Student

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)}.