the line 3y = 4x - 15 intersects thje curve 8x^2-27^2 =45 at points a and b find points a and b

To find the points \( A \) and \( B \) where the line \( 3y = 4x - 15 \) intersects the curve \( 8x^2 - 27y^2 = 45 \), we need to solve the system of equations. Let's go step by step:

### Step 1: Express \( y \) from the line equation
The equation of the line is:

\[
3y = 4x - 15
\]

Solve for \( y \):

\[
y = \frac{4x - 15}{3}
\]

### Step 2: Substitute \( y \) in the curve equation
Now substitute \( y = \frac{4x - 15}{3} \) into the curve equation:

\[
8x^2 - 27y^2 = 45
\]

Substituting for \( y \):

\[
8x^2 - 27\left(\frac{4x - 15}{3}\right)^2 = 45
\]

### Step 3: Simplify the equation
First, square the expression for \( y \):

\[
\left(\frac{4x - 15}{3}\right)^2 = \frac{(4x - 15)^2}{9}
\]

So the equation becomes:

\[
8x^2 - 27 \cdot \frac{(4x - 15)^2}{9} = 45
\]

Simplify the fraction:

\[
8x^2 - 3 \cdot (4x - 15)^2 = 45
\]

### Step 4: Expand \( (4x - 15)^2 \)
Expand \( (4x - 15)^2 \):

\[
(4x - 15)^2 = 16x^2 - 120x + 225
\]

Substitute this into the equation:

\[
8x^2 - 3(16x^2 - 120x + 225) = 45
\]

### Step 5: Simplify the equation
Distribute the 3:

\[
8x^2 - (48x^2 - 360x + 675) = 45
\]

Simplify further:

\[
8x^2 - 48x^2 + 360x - 675 = 45
\]

Combine like terms:

\[
-40x^2 + 360x - 675 = 45
\]

Subtract 45 from both sides:

\[
-40x^2 + 360x - 720 = 0
\]

### Step 6: Solve the quadratic equation
Divide the equation by -40 to simplify:

\[
x^2 - 9x + 18 = 0
\]

Now solve this quadratic equation using the quadratic formula:

\[
x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(18)}}{2(1)}
\]

\[
x = \frac{9 \pm \sqrt{81 - 72}}{2}
\]

\[
x = \frac{9 \pm \sqrt{9}}{2}
\]

\[
x = \frac{9 \pm 3}{2}
\]

So the two solutions for \( x \) are:

\[
x = \frac{9 + 3}{2} = 6 \quad \text{or} \quad x = \frac{9 - 3}{2} = 3
\]

### Step 7: Find the corresponding \( y \) values
Now substitute these values of \( x \) into the equation for \( y = \frac{4x - 15}{3} \):

- For \( x = 6 \):

\[
y = \frac{4(6) - 15}{3} = \frac{24 - 15}{3} = \frac{9}{3} = 3
\]

- For \( x = 3 \):

\[
y = \frac{4(3) - 15}{3} = \frac{12 - 15}{3} = \frac{-3}{3} = -1
\]

### Step 8: Conclusion
The points of intersection are:

\[
A(6, 3) \quad \text{and} \quad B(3, -1)
\]

Would you like any further explanation or more details?

Here are 5 related questions for further practice:
1. How do you solve systems of equations with a line and a circle?
2. Can you find the distance between points \( A(6, 3) \) and \( B(3, -1) \)?
3. What happens if the line does not intersect the curve?
4. How do you solve a system with a line and a parabola?
5. How does the discriminant help determine the nature of the roots in quadratic equations?

**Tip:** For quadratic equations, always check the discriminant (\(b^2 - 4ac\)) to determine whether the equation has real, imaginary, or repeated roots.

The line 3y = 4x - 15 intersects the curve 8x^2 - 27y^2 = 45 at points A and B. Find points A and B.

Solve the intersection problem between the line 3y = 4x - 15 and the curve 8x^2 - 27y^2 = 45 with a comprehensive step-by-step guide. Learn how to substitute and solve systems of equations involving lines and curves, applying quadratic formulas. Ideal for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation