Solving Complex Systems Of Equations

Welcome to our deep dive into solving systems of equations! Today, we're tackling a particularly interesting challenge: a system involving both a parabola and an ellipse. This kind of problem can seem daunting at first glance, but by breaking it down step-by-step, we can unravel its secrets and find all the possible solutions. Let's get started on our mathematical journey to understand how to find the points where these two distinct curves intersect.

Understanding the Equations

Our system of equations is given by:

1. $3x^2+18x-8y+8=0$
2. $3x^2+4y^2+18x-40y+36=0$

Let's first get a feel for what these equations represent graphically. The first equation, $3x^2+18x-8y+8=0$, can be rearranged to isolate $y$. Doing so, we get $8y = 3x^2 + 18x + 8$, or $y = \frac{3}{8}x^2 + \frac{18}{8}x + \frac{8}{8}$, which simplifies to $y = \frac{3}{8}x^2 + \frac{9}{4}x + 1$. This is the equation of a parabola that opens upwards. Its vertex and axis of symmetry can be found using standard techniques, but for solving the system, knowing it's a parabola is key.

The second equation, $3x^2+4y^2+18x-40y+36=0$, is a bit more complex. Notice the presence of both $x^2$ and $y^2$ terms with different coefficients. This indicates that it's the equation of an ellipse. To see this more clearly, we can complete the square for both the $x$ and $y$ terms. First, let's group the terms: $(3x^2+18x) + (4y^2-40y) + 36 = 0$.

Now, complete the square for the $x$ terms: $3(x^2+6x) = 3(x^2+6x+9-9) = 3(x+3)^2 - 27$.

Next, complete the square for the $y$ terms: $4(y^2-10y) = 4(y^2-10y+25-25) = 4(y-5)^2 - 100$.

Substitute these back into the equation: $[3(x+3)^2 - 27] + [4(y-5)^2 - 100] + 36 = 0$.

Simplify: $3(x+3)^2 + 4(y-5)^2 - 27 - 100 + 36 = 0$.

$3(x+3)^2 + 4(y-5)^2 - 91 = 0$.

$3(x+3)^2 + 4(y-5)^2 = 91$.

Dividing by 91 to get the standard form of an ellipse: $\frac{(x+3)^2}{91/3} + \frac{(y-5)^2}{91/4} = 1$. This confirms it's an ellipse centered at $(-3, 5)$. The different denominators, $91/3 \approx 30.33$ and $91/4 = 22.75$, tell us about the lengths of its semi-axes. The intersection points of the parabola and the ellipse are the solutions to our system.

Strategies for Solving the System

We have two equations:

1. $3x^2+18x-8y+8=0$
2. $3x^2+4y^2+18x-40y+36=0$

Notice that both equations share the terms $3x^2$ and $18x$. This commonality is our biggest clue for an efficient solution strategy. We can use the method of elimination or substitution. Let's try elimination first, as it seems promising here.

If we subtract the first equation from the second equation, we can eliminate the $3x^2$ and $18x$ terms entirely. Let's write it out:

$(3x^2+4y^2+18x-40y+36) - (3x^2+18x-8y+8) = 0 - 0$

$3x^2+4y^2+18x-40y+36 - 3x^2-18x+8y-8 = 0$

Combine like terms:

$(3x^2 - 3x^2) + 4y^2 + (18x - 18x) + (-40y + 8y) + (36 - 8) = 0$

This simplifies beautifully to:

$4y^2 - 32y + 28 = 0$

We can divide this entire equation by 4 to make it even simpler:

$y^2 - 8y + 7 = 0$

This is a quadratic equation in $y$. We can solve it by factoring. We are looking for two numbers that multiply to 7 and add up to -8. These numbers are -1 and -7. So, the equation factors as:

$(y-1)(y-7) = 0$

This gives us two possible values for $y$: $y=1$ or $y=7$. These are the $y$-coordinates of the intersection points. Now, we need to find the corresponding $x$-coordinates for each of these $y$ values.

Finding the Corresponding x-coordinates

We will use the first equation (the parabola's equation) to find the $x$-values because it's simpler to solve for $x$ in terms of $y$ (or substitute $y$ into it directly). The first equation is $3x^2+18x-8y+8=0$. Let's rearrange it to make substitution easier:

$3x^2+18x = 8y - 8$

Now, we'll consider each $y$ value we found:

Case 1: $y=1$

Substitute $y=1$ into the rearranged equation:

$3x^2+18x = 8(1) - 8$

$3x^2+18x = 8 - 8$

$3x^2+18x = 0$

We can factor out $3x$ from this equation:

$3x(x+6) = 0$

This gives us two possible values for $x$: $3x=0$ or $x+6=0$. Therefore, $x=0$ or $x=-6$.

So, for $y=1$, we have two potential solutions: $(0, 1)$ and $(-6, 1)$.

Case 2: $y=7$

Substitute $y=7$ into the rearranged equation:

$3x^2+18x = 8(7) - 8$

$3x^2+18x = 56 - 8$

$3x^2+18x = 48$

To solve this quadratic equation for $x$, let's move all terms to one side:

$3x^2+18x - 48 = 0$

We can divide the entire equation by 3:

$x^2+6x - 16 = 0$

Now, we factor this quadratic. We need two numbers that multiply to -16 and add up to 6. These numbers are 8 and -2. So, the equation factors as:

$(x+8)(x-2) = 0$

This gives us two possible values for $x$: $x+8=0$ or $x-2=0$. Therefore, $x=-8$ or $x=2$.

So, for $y=7$, we have two potential solutions: $(-8, 7)$ and $(2, 7)$.

Verifying the Solutions

We have found four potential solutions: $(0, 1)$, $(-6, 1)$, $(-8, 7)$, and $(2, 7)$. It's crucial to verify these solutions in both original equations to ensure they are correct and not extraneous. However, since we derived the $y$ values by eliminating $x$ and then used the simpler equation to find $x$, these points should lie on both curves. Let's pick one point to double-check, say $(0, 1)$, in both original equations.

Equation 1: $3x^2+18x-8y+8=0$

Substitute $(0, 1)$: $3(0)^2 + 18(0) - 8(1) + 8 = 0 + 0 - 8 + 8 = 0$. This works.

Equation 2: $3x^2+4y^2+18x-40y+36=0$

Substitute $(0, 1)$: $3(0)^2 + 4(1)^2 + 18(0) - 40(1) + 36 = 0 + 4(1) + 0 - 40 + 36 = 4 - 40 + 36 = 0$. This also works.

Let's verify another point, say $(-8, 7)$.

Equation 1: $3x^2+18x-8y+8=0$

Substitute $(-8, 7)$: $3(-8)^2 + 18(-8) - 8(7) + 8 = 3(64) - 144 - 56 + 8 = 192 - 144 - 56 + 8 = 48 - 56 + 8 = -8 + 8 = 0$. This works.

Equation 2: $3x^2+4y^2+18x-40y+36=0$

Substitute $(-8, 7)$: $3(-8)^2 + 4(7)^2 + 18(-8) - 40(7) + 36 = 3(64) + 4(49) - 144 - 280 + 36 = 192 + 196 - 144 - 280 + 36 = 388 - 144 - 280 + 36 = 244 - 280 + 36 = -36 + 36 = 0$. This works too.

Given the way we solved it, all four points derived from the values of $y$ should be valid solutions. The options provided are:

A. (-5.5, 0) B. (-0.48, 0) C. (-8, 7) D. (-6, 1) E. (0, 1) F. (2, 7) G. (0, 9) H. The system has no solutions. I. The system has infinitely many

Comparing our found solutions with the options, we see that:

• $(-8, 7)$ matches option C.
• $(-6, 1)$ matches option D.
• $(0, 1)$ matches option E.
• $(2, 7)$ matches option F.

These are all the intersection points. Therefore, the system has exactly these four solutions.

Conclusion

Solving systems of equations, especially those involving different conic sections like parabolas and ellipses, requires a methodical approach. By recognizing common terms and employing elimination, we simplified the system considerably, reducing it to a quadratic equation in $y$. Solving for $y$ gave us the possible $y$-coordinates of the intersection points. Substituting these $y$-values back into one of the original equations allowed us to find the corresponding $x$-coordinates. The process yielded four distinct solutions: $(-8, 7)$, $(-6, 1)$, $(0, 1)$, and $(2, 7)$. These are the points where the parabola and the ellipse intersect. It's always a good practice to verify these solutions in both original equations to confirm their validity.

For further exploration into the fascinating world of conic sections and solving systems of equations, you can visit Khan Academy for excellent tutorials and practice problems. Understanding the graphical representation of these equations can also significantly aid in visualizing the solutions. You might also find resources on Wolfram MathWorld incredibly helpful for in-depth mathematical definitions and theorems related to conic sections and algebraic geometry.