Problem 50 - Algebra A Combined Function

Question

The graph of each equation is an ellipse. Determine which distance is longer, the distance between the \(x\) -intercepts or the distance between the y-intercepts. How much longer? $$ \frac{x^{2}}{100}+\frac{y^{2}}{49}=1 $$

Step-by-Step Solution

Answer

The distance between the x-intercepts is longer by 6 units.

Step 1: Identify the Equation of the Ellipse

The given equation of the ellipse is \(\frac{x^2}{100} + \frac{y^2}{49} = 1\). This represents an ellipse centered at the origin, with the semi-major axis along the x-axis and the semi-minor axis along the y-axis.

Step 2: Calculate the x-Intercepts

To find the x-intercepts, set \(y = 0\). The equation becomes \(\frac{x^2}{100} = 1\). Solving for \(x\), we get \(x^2 = 100\), so \(x = \pm 10\). Therefore, the x-intercepts are \((10, 0)\) and \((-10, 0)\).

Step 3: Calculate the y-Intercepts

To find the y-intercepts, set \(x = 0\). The equation becomes \(\frac{y^2}{49} = 1\). Solving for \(y\), we get \(y^2 = 49\), so \(y = \pm 7\). Therefore, the y-intercepts are \((0, 7)\) and \((0, -7)\).

Step 4: Calculate the Distance Between x-Intercepts

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). For the x-intercepts \((10, 0)\) and \((-10, 0)\), the distance is calculated as: \[\sqrt{(10 - (-10))^2 + (0 - 0)^2} = \sqrt{20^2} = 20\].

Step 5: Calculate the Distance Between y-Intercepts

For the y-intercepts \((0, 7)\) and \((0, -7)\), the distance is given by: \[\sqrt{(0 - 0)^2 + (7 - (-7))^2} = \sqrt{14^2} = 14\].

Step 6: Compare Distances and Find the Difference

The distance between the x-intercepts is 20, while the distance between the y-intercepts is 14. The distance between the x-intercepts is longer by \(20 - 14 = 6\).

Key Concepts

x-intercepts

When dealing with the equation of an ellipse, identifying the x-intercepts can be crucial. X-intercepts, where the ellipse crosses the x-axis, occur when the y-coordinate is zero. We set up the ellipse equation \(\frac{x^2}{100} + \frac{y^2}{49} = 1\), and replace y with zero, simplifying to \(\frac{x^2}{100} = 1\). By solving, we find \(x^2 = 100\), which gives \(x = \pm 10\). Thus, the x-intercepts are at the points \( (10, 0) \) and \( (-10, 0) \). These help us understand how wide the ellipse stretches along the x-axis. By following these steps, you can easily find the x-intercepts for similar ellipse equations.

y-intercepts

Y-intercepts are points where the ellipse cuts across the y-axis. To find them, we start with the elliptical equation \(\frac{x^2}{100} + \frac{y^2}{49} = 1\) and set x to zero, transforming the equation into \(\frac{y^2}{49} = 1\). Solving for y, we get \(y^2 = 49\), which concludes in \(y = \pm 7\). This indicates the y-intercepts are located at \( (0, 7) \) and \( (0, -7) \). These points show the maximum reach of the ellipse along the y-axis, illustrating its vertical span. With these steps, solving for y-intercepts in other ellipse problems can be straightforward.

distance formula

The distance formula is a powerful tool used to determine the space between two points on a plane. It is expressed as \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\. For understanding, let's apply it to the x-intercepts of our ellipse, which are \( (10, 0) \) and \( (-10, 0) \). Plugging these into the formula gives \sqrt{(10 - (-10))^2 + (0 - 0)^2}\, simplifying to \sqrt{20^2} = 20\. For the y-intercepts, \( (0, 7) \) and \( (0, -7) \), the calculation is \sqrt{(0-0)^2 + (7 - (-7))^2}\, which simplifies to \sqrt{14^2} = 14\. Every time you need to find a straight-line distance, this formula proves to be very handy.

semi-major axis

The semi-major axis is the longest radius of an ellipse, extending from the center to its furthest edge along the major direction. In the equation \(\frac{x^2}{100} + \frac{y^2}{49} = 1\), we see that the semi-major axis aligns with the x-axis since \(100 > 49\). The distance from the center (0,0) to one end is \rac{\sqrt{100}}{1} = 10\ units. This provides the full stretch from the center to the outer part of the ellipse along the x-axis, showing us the greater reach compared to the semi-minor axis. Establishing which axis is major helps understand how the ellipse elongates in one direction.