Problem 87 - Algebra and Trigonometry

Question

determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. The graph of \(\frac{x^{2}}{9}-\frac{y^{2}}{4}=1\) does not intersect the line \(y=-\frac{2}{3} x\)

Step-by-Step Solution

Answer

The statement is false. The graph of the given hyperbola equation does intersect with the line represented by the given linear equation. The necessary change is to remove the word 'not' from the original statement.

Step 1: Set the equations equal to each other

To find the intersection points, the y-values of both equations must be equal. So, set \(-\frac{2}{3} x\) equal to the y-value found in the equation of the hyperbola. Replace \(y\) in the first equation with \(-\frac{2}{3} x\). This gives us: \(\frac{x^{2}}{9}-\frac{(-\frac{2}{3} x)^{2}}{4}=1\)

Step 2: Simplify the equation

Simplify the new equation by performing the calculation inside the bracket and further rearranging the equation to solve for 'x'. That gives us the following equation: \(x^{2}-\frac{4}{9}x^{2}=9\)

Step 3: Combine like terms

Combine like terms and simplify the equation further: \(x^{2}-\frac{4}{9}x^{2}=9 => \frac{5}{9}x^{2}=9\)

Step 4: Solve for 'x'

Find the value of 'x' by dividing 9 by \(\frac{5}{9}\), which results in \(x^{2}=\frac{162}{5}\)

Step 5: Square root of 'x'

Since \(x^{2}=\frac{162}{5}\), x is equal to the square root of \(\frac{162}{5}\). Since it yields real positive and negative values, it means that the line and the hyperbola intersect at two points.

Key Concepts

Solving Systems of Equations

When you encounter two algebraic expressions that represent different geometric shapes, like a hyperbola and a line, finding their points of intersection involves solving a system of equations. Solving these systems is a fundamental skill in algebra and is essential for understanding more complex mathematical concepts.

To solve the system, you equate the expressions for 'y' from both the hyperbola and the line equations, as done in Step 1 of the provided solution. This common value of 'y' is the point at which they intersect on the y-axis. Once you've equated the equations, you have essentially set up a single-variable equation that you can solve for 'x'. This method is very useful, as it can be applied to linear, quadratic, or even more complex systems.

Graph of Hyperbola

Understanding the graph of a hyperbola is crucial when solving these types of problems. A hyperbola is a set of points found by subtracting distances from two fixed points, called foci. The standard form of a hyperbola centered at the origin with horizontal transverse axis is \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\) and with vertical transverse axis is \(\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1\).

In this exercise, the equation \(\frac{x^{2}}{9}-\frac{y^{2}}{4}=1\) represents a hyperbola centered at the origin with a horizontal transverse axis. The values under the \(x^{2}\) and \(y^{2}\) terms indicate the squares of the transverse and conjugate axis lengths, which can be used to sketch the graph of the hyperbola and determine its orientation.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is a process that often makes it possible to solve equations that would otherwise be too complex to handle. In our example, once we substitute the expression of 'y' from the line into the hyperbola's equation, we're left with a more complex algebraic expression that requires simplification (Step 2).

To simplify, we perform arithmetic operations and combine like terms. This process can include squaring binomials, distributing multiplicative factors, adding or subtracting similar terms, and isolating the variable you're solving for by addition, subtraction, multiplication, or division. In essence, your goal is to rewrite the expression in an equivalent form that is easier to interpret and solve, as seen in Steps 3 and 4.

Finding Intersection Points

Finding intersection points is essentially the end goal when solving systems of equations involving a hyperbola and a line. After simplifying our algebraic expression, we determine the 'x' values where the hyperbola and line intersect by solving for 'x' (Step 4).

These 'x' values are then used to find the corresponding 'y' values by plugging them back into either the original equation of the hyperbola or the line, which finally gives us the coordinates of the intersection points. If we find real number solutions for 'x', then it confirms that intersections exist, and the number of intersection points depends on the number of unique solutions we obtain. In this case, we found that the line intersects the hyperbola at two points, because the equation has two real solutions for 'x'.