Question

Write each general equation in standard form. Find the center, foci, vertices, and the lengths \(a\) and \(b\) of the semi axes, and make a graph.$$49 x^{2}+81 y^{2}+294 x+810 y-1503=0$$

Short answer

The standard form is \(\frac{(x + 3)^2}{81} + \frac{(y + 5)^2}{49} = 1\). The center is \((-3, -5)\), the semi-major axis length \

Step-by-step solution

Write in Standard Form

To write the given equation in standard form, complete the squares for the x-terms and the y-terms. Given equation: \(49x^{2} + 294x + 81y^{2} + 810y - 1503 = 0\).First, group the x-terms together and the y-terms together: \(49(x^{2} + 6x) + 81(y^{2} + 10y) - 1503 = 0\).Find the terms to complete the squares: \((6/2)^2 = 9\) and \((10/2)^2 = 25\).Rewrite the equation as \(49(x^2 + 6x + 9) + 81(y^2 + 10y + 25) - 1503 - 49\cdot9 - 81\cdot25 = 0\). Simplify to get the standard form: \(49(x + 3)^2 + 81(y + 5)^2 = 1503 + 441 + 2025\), which simplifies to \(49(x + 3)^2 + 81(y + 5)^2 = 3969\). Finally, divide both sides by 3969 to get \(\frac{(x + 3)^2}{81} + \frac{(y + 5)^2}{49} = 1\).

Identify the Elements

From the standard form equation \(\frac{(x + 3)^2}{81} + \frac{(y + 5)^2}{49} = 1\), we can identify the following elements:The center \((h, k)\) is the point \((-3, -5)\).The semi-major axis 'a' corresponds to the larger denominator under the x or y term, which is \(\frac{81}{49} = 9 \rightarrow a = 9\).The semi-minor axis 'b' corresponds to the smaller denominator under the x or y term, which is \(b = 7\).The vertices are located \(a\) units away from the center along the major axis, so they are at points \((-3 \pm 9, -5) = (-12, -5)\) and \((6, -5)\).The foci are found by calculating \(c = \sqrt{a^2 - b^2}\), which is \(c = \sqrt{81 - 49} = \sqrt{32} = 4\sqrt{2}\). The foci are at points \((-3 \pm 4\sqrt{2}, -5)\).

Make a Graph

To graph the ellipse, plot the center at \((-3, -5)\). Draw the major axis (horizontal) with lengths of \(9\) units on either side of the center, and the minor axis (vertical) with lengths of \(7\) units. Mark the vertices at \((-12, -5)\) and \((6, -5)\), as well as marking the co-vertices at \((-3, -5 \pm 7) = (-3, -12)\) and \((-3, 2)\). Lastly, plot the foci at points \((-3 \pm 4\sqrt{2}, -5)\). Sketch the ellipse through the vertices and co-vertices. Make sure the curve is closer to the center at the vertices along the minor axis and further from the center along the major axis.

Key concepts

Completing the Square

Completing the square is an algebraic technique used to convert quadratic equations into their standard form. This process is essential when dealing with ellipses and other conic sections because it allows us to clearly identify the properties of the shape, such as the center, vertices, foci, and axes lengths.

To complete the square for the equation 'ax^2 + bx + c', you work with the 'x' terms: factor out 'a' from 'x^2 + bx', and add a new term to turn this into a perfect square trinomial. The term to be added is \(\left(\frac{b}{2a}\right)^2\). Once the perfect square is formed, remember to balance the equation by subtracting the added value after multiplying it back by the original 'a'.

This technique is used in two dimensions for ellipses, dealing with both 'x' and 'y' terms separately. After manipulating the quadratic terms for both 'x' and 'y', you will get the equation of an ellipse in the standard form. This form is useful because it immediately shows the lengths of the semi-major axis 'a' and the semi-minor axis 'b', from which you can derive other properties of the ellipse.

Ellipse Elements

Understanding ellipse elements is crucial when analyzing its equation in standard form. The basic elements of an ellipse include the center, vertices, foci, lengths of the semi-major axis (denoted as 'a') and the semi-minor axis (denoted as 'b'). The center \(\text{(h, k)}\) of the ellipse is the midpoint of the major and minor axes and serves as a reference point for the entire curve.

Vertices are the points on the ellipse furthest from the center along the major axis, and co-vertices are points furthest along the minor axis. The foci are two fixed points located along the major axis, from which the sum of distances to any point on the ellipse is constant. The distance from the center to each focus, 'c', is found using the equation \(c = \sqrt{a^2 - b^2}\).

These elements are directly found from the standard form equation of the ellipse, \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\). By identifying these elements, we can gather enough information to accurately sketch the ellipse and understand its geometry.

Graphing Ellipses

Graphing ellipses requires a systematic approach starting from its standard form equation. First, you should plot the center of the ellipse on a coordinate system. The center is determined from the standard form by identifying the values of 'h' and 'k' in the denominators' expressions.

Next, draw the major and minor axes through the center. The lengths of these axes are determined by '2a' and '2b', respectively, which represents the distance from the center to the vertices (major axis) and to the co-vertices (minor axis). Then, the vertices and co-vertices are marked at the appropriate distances along the respective axes.

Plotting the Foci

To find the foci, calculate the distance 'c' from the center using the equation \(c = \sqrt{a^2 - b^2}\). Mark the two foci positions on the major axis symmetrically from the center. The final step is to sketch the ellipse. Make sure the ellipse passes through the vertices and co-vertices, maintaining a smooth, curved shape that is elongated along the major axis and compressed along the minor axis.

It's helpful to check your graph by ensuring that the distance from one focus, through any point on the ellipse, to the other focus sums to '2a'. This is a defining property of ellipses and reflects their geometric foundation. By graphing the ellipse with precision, the visual representation helps students grasp the relationship between the algebraic equation and its geometric counterpart.