Question

The line 5x – 5y = 2 intersects the curve x2y – 5x + y + 2 = 0 at

three points.
(a) Find the coordinates of the points of intersection
(b) Find the gradient of the curve at each of the points on intersection.​

Answer 1

(a) The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve at each point of intersection are;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28

Explanation:

The equations of the lines are;

5·x - 5·y = 2......(1)

x²·y - 5·x + y + 2 = 0.......(2)

Making y the subject of equation (1) gives;

5·y = 5·x - 2

y = (5·x - 2)/5

Making y the subject of equation (2) gives;

y·(x² + 1) - 5·x + 2 = 0

y = (5·x - 2)/(x² + 1)

Therefore, at the point the two lines intersect their coordinates are equal thus we have;

y = (5·x - 2)/5 = y = (5·x - 2)/(x² + 1)

Which gives;

`(5 \cdot x - 2)/(5) = (5 \cdot x - 2)/(x^2 + 1)`

Therefore, 5 = x² + 1

x² = 5 - 1 = 4

x = √4 = 2

Which is an indication that the x-coordinate is equal to 2

The y-coordinate is therefore;

y = (5·x - 2)/5 = (5 × 2 - 2)/5 = 8/5

The coordinates of the points of intersection = (2, 8/5}

Cross multiplying the following equation

Substituting the value for y in equation (2) with (5·x - 2)/5 gives;

`(5 \cdot x^3 - 2 \cdot x^2 - 20 \cdot x + 8)/(5) = 0`

Therefore;

5·x³ - 2·x² - 20·x + 8 = 0

(x - 2)×(5·x² - b·x + c) = 5·x³ - 2·x² - 20·x + 8

Therefore, we have;

x²·b - 2·x·b -x·c + 2·c -5·x³ + 10·x²

5·x³ - 10·x² - x²·b + 2·x·b + x·c - 2·c = 5·x³ - 2·x² - 20·x + 8

∴ c = 8/(-2) = -4

2·b + c = - 20

b = -16/2 = -8

Therefore;

(x - 2)×(5·x² - b·x + c) = (x - 2)×(5·x² + 8·x - 4)

(x - 2)×(5·x² + 8·x - 4) = 0

5·x² + 8·x - 4 = 0

x² + 8/5·x - 4/5 = 0

(x + 4/5)² - (4/5)² - 4/5 = 0

(x + 4/5)² = 36/25

x + 4/5 = ±6/5

x = 6/5 - 4/5 = 2/5 or -6/5 - 4/5 = -2

Hence the three x-coordinates are

x = 2, x = - 2, and x = 2/5

The y-coordinates are derived from y = (5·x - 2)/5 as y = 8/5, y = -12/5, and y = y = 0

The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve,
`\frac{\mathrm{d} y}{\mathrm{d} x}`, is given by the differentiation of the equation of the curve, x²·y - 5·x + y + 2 = 0 which is the same as y = (5·x - 2)/(x² + 1)

Therefore, we have;

`\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\mathrm{d} \left ((5 \cdot x - 2)/(x^2 + 1) \right )}{\mathrm{d} x} = (5\cdot \left ( x^(2) +1\right )-\left ( 5\cdot x-2 \right )\cdot 2\cdot x)/(\left (x^2 + 1 ^(2) \right ))`.......(3)

Which gives by plugging in the value of x in the slope equation;

At x = -2,
`\frac{\mathrm{d} y}{\mathrm{d} x}` = -0.92

At x = 2/5,
`\frac{\mathrm{d} y}{\mathrm{d} x}` = 4.3

At x = 2,
`\frac{\mathrm{d} y}{\mathrm{d} x}` = -0.28

Therefore;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28.