Question

The line 2x + 3y = 8 meets the curve 2x^2 + 3y^2 =110 at the points A and B. Find the coordinates of A and B.

extra: This is the first time I’ve seen this type of question without a diagram. Pls help!

Answer 1

The coordinates are A. (-3.8, 5.2) and B. (7, -2).

Explanation:

At A and B the coordinates satisfy both equations so we solve simultaneously using substitution:

2x^2 + 3y^2 = 110

2x + 3y = 8

From the above equation

x = (8 - 3y)/2

So substituting for x in the first equation:

2 [(8-3y)/2]^2 + 3y^2 = 110

(8-3y)^2 / 2 + 3y^2 = 110

Multiply through by 2:

(8 - 3y)^2 + 6y^2 = 220

64 + 9y^2 - 48y + 6y^2 - 220 = 0

15y^2 - 48y - 156 = 0

3(5y^2 - 16y - 52) = 0

(5y - 26)(y + 2) = 0

y = 26/5, 2

y = 5.2, -2.

So substituting in 2x + 3y = 8

When y = -2:

2x - 6 = 8

2x = 14

x = 7.

When y = 5.2

2x + 3(5.2) = 8

2x = 8 - 15.6 = -7.6

x = -3.8.

Answer 2

Answer:The cordinates of A and B are (-3.8, 5.2) and (7, -2).

Step-by-step explanation:From (1), we get x = (8 - 3y) / 2 ––(3)

Substitute (3) into (2),

2 [(8 - 3y) / 2]² + 3y² = 110

(8 - 3y)² + 3y² = 220

15y² - 48y - 156 = 0

Therefore y = 5.2, then x = (8 - 3×5.2) / 2 = -3.8

OR y = -2, then x = [8 - 3×(-2)] / 2 = 7