line 4y =4x-14 intersects curve 8x^2-27y^2=45 at points a and b find a and b

To find the points \(A\) and \(B\) where the line intersects the curve, we'll solve the system of equations:

1. The equation of the line is:
   \[
   4y = 4x - 14 \quad \text{or} \quad y = x - \frac{7}{2}
   \]
   
2. The equation of the curve is:
   \[
   8x^2 - 27y^2 = 45
   \]

### Step 1: Substitute the line equation into the curve equation
We substitute \(y = x - \frac{7}{2}\) into the curve equation \(8x^2 - 27y^2 = 45\).

\[
8x^2 - 27\left(x - \frac{7}{2}\right)^2 = 45
\]

### Step 2: Expand and simplify
First, expand \(\left(x - \frac{7}{2}\right)^2\):
\[
\left(x - \frac{7}{2}\right)^2 = x^2 - 7x + \frac{49}{4}
\]

Now substitute this back into the equation:
\[
8x^2 - 27\left(x^2 - 7x + \frac{49}{4}\right) = 45
\]

Distribute the \(-27\):
\[
8x^2 - 27x^2 + 189x - \frac{1323}{4} = 45
\]

Multiply everything by 4 to eliminate the fraction:
\[
32x^2 - 108x^2 + 756x - 1323 = 180
\]

Simplify the terms:
\[
-76x^2 + 756x - 1323 = 180
\]

Move everything to one side:
\[
-76x^2 + 756x - 1503 = 0
\]

### Step 3: Solve the quadratic equation
Divide the entire equation by \(-1\) to make it easier to solve:
\[
76x^2 - 756x + 1503 = 0
\]

We will now use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 76\), \(b = -756\), and \(c = 1503\).

Let's calculate the discriminant \( \Delta = b^2 - 4ac \):

\[
\Delta = (-756)^2 - 4(76)(1503)
\]

Now, solving for \(x\):The solutions for \(x\) are:

\[
x_1 \approx 7.20 \quad \text{and} \quad x_2 \approx 2.75
\]

### Step 4: Find the corresponding \(y\) values
Using the line equation \(y = x - \frac{7}{2}\), we can find the corresponding \(y\) values for each \(x\):

- For \(x_1 = 7.20\):
  \[
  y_1 = 7.20 - \frac{7}{2} = 7.20 - 3.5 = 3.70
  \]

- For \(x_2 = 2.75\):
  \[
  y_2 = 2.75 - \frac{7}{2} = 2.75 - 3.5 = -0.75
  \]

### Step 5: Conclusion
The points of intersection are:
\[
A(7.20, 3.70) \quad \text{and} \quad B(2.75, -0.75)
\]

Would you like further clarification on any step?

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Here are 5 related questions:
1. How do you derive the quadratic formula?
2. Can this method be applied to non-linear systems involving trigonometric functions?
3. What happens if the discriminant is negative?
4. How do you find the tangent line to a curve at a point?
5. How do we calculate the area between two intersecting curves?

**Tip:** Always check the discriminant to determine the nature of the roots before solving a quadratic equation.

line 4y = 4x - 14 intersects curve 8x^2 - 27y^2 = 45 at points A and B. Find A and B.

Learn how to find the points of intersection between the line 4y = 4x - 14 and the curve 8x^2 - 27y^2 = 45. This problem involves solving a quadratic equation derived from substituting the line equation into the curve equation. Suitable for students in Grades 10-12.

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