(a) Finding the Coordinates of Points P and Q

The problem involves finding the points of intersection between the curve $(y + 1)^2 - x^2 = 8$ and the straight line $3x + 2y = -5$.

Step 1: Express $y$ from the line equation

We start by solving the line equation for $y$:

$3x + 2y = -5 \implies 2y = -3x - 5 \implies y = -\frac{3}{2}x - \frac{5}{2}$

Step 2: Substitute into the curve equation

Next, substitute this expression for $y$ into the curve equation $(y + 1)^2 - x^2 = 8$:

$\left(-\frac{3}{2}x - \frac{5}{2} + 1\right)^2 - x^2 = 8$

Simplifying the expression inside the square:

$y + 1 = -\frac{3}{2}x - \frac{3}{2} \implies \left(-\frac{3}{2}(x + 1)\right)^2 - x^2 = 8$

Simplifying the square:

$\left(\frac{9}{4}(x + 1)^2\right) - x^2 = 8$

Expanding and simplifying:

$\frac{9}{4}(x^2 + 2x + 1) - x^2 = 8 \implies \frac{9}{4}x^2 + \frac{9}{2}x + \frac{9}{4} - x^2 = 8$

Multiply by 4 to eliminate the fraction:

$9x^2 + 18x + 9 - 4x^2 = 32$

$5x^2 + 18x - 23 = 0$

Step 3: Solve the quadratic equation

We solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 5$, $b = 18$, and $c = -23$:

$x = \frac{-18 \pm \sqrt{18^2 - 4(5)(-23)}}{2(5)} = \frac{-18 \pm \sqrt{324 + 460}}{10} = \frac{-18 \pm \sqrt{784}}{10} = \frac{-18 \pm 28}{10}$

This gives us two possible values for $x$:

$x_1 = 1 \quad \text{and} \quad x_2 = -4.6$

Step 4: Find corresponding $y$ values

Substitute these $x$ values back into $y = -\frac{3}{2}x - \frac{5}{2}$:

For $x_1 = 1$:

$y_1 = -\frac{3}{2}(1) - \frac{5}{2} = -\frac{3}{2} - \frac{5}{2} = -\frac{8}{2} = -4$

For $x_2 = -4.6$:

$y_2 = -\frac{3}{2}(-4.6) - \frac{5}{2} = \frac{13.8}{2} - \frac{5}{2} = \frac{8.8}{2} = 4.4$

So, the coordinates of $P$ and $Q$ are:

$\text{Coordinates of } P = (1, -4) \quad \text{and} \quad Q = (-4.6, 4.4)$

(b) Finding the Coordinates of Point R

Now, we find the intersection of the lines $3x + 2y = -5$ and $y - 2x = 1$.

Step 1: Express $y$ from the second line equation

$y = 2x + 1$

Step 2: Substitute into the first line equation

Substitute $y = 2x + 1$ into $3x + 2y = -5$:

$3x + 2(2x + 1) = -5 \implies 3x + 4x + 2 = -5 \implies 7x = -7 \implies x = -1$

Step 3: Find the corresponding $y$ value

Substitute $x = -1$ into $y = 2x + 1$:

$y = 2(-1) + 1 = -2 + 1 = -1$

Thus, the coordinates of $R$ are:

$\text{Coordinates of } R = (-1, -1)$

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Summary of Results:

• Coordinates of point P = $(1, -4)$
• Coordinates of point Q = $(-4.6, 4.4)$
• Coordinates of point R = $(-1, -1)$

Would you like to see any specific details or have any questions on this solution?

Related Questions:

1. How do you determine the number of intersection points between a line and a curve?
2. Can a quadratic equation have complex solutions, and if so, what does that mean geometrically?
3. How does the discriminant of a quadratic equation affect the nature of its roots?
4. What are some methods to solve a system of linear equations geometrically?
5. How can you interpret the intersection of two conic sections in a plane?

Tip:

When solving systems of equations involving curves, substitution is a powerful method, but always simplify expressions carefully to avoid errors in the subsequent algebraic steps.