Problem 50 - Algebra and Trigonometry

Question

In Exercises \(37-50,\) graph each ellipse and give the location of its foci. $$ 36(x+4)^{2}+(y+3)^{2}=36 $$

Step-by-Step Solution

Answer

The center of the ellipse is \((-4, -3)\), the values of semi-major axis (b) and semi-minor axis (a) are \(6\) and \(1\), respectively. The foci lie on \((-4, -3 + \sqrt{35})\) and \((-4, -3 - \sqrt{35})\).

Step 1: Find the Center of the Ellipse

The ellipse equation is in the form \((x-h)^{2}/a^{2} + (y-k)^{2}/b^{2} = 1\), where (h,k) is the center of the ellipse. Comparing our equation \(36(x-(-4))^{2}+(y-(-3))^{2}=36\) with the standard form, we find that the center of the ellipse is \((-4,-3)\).

Step 2: Identify the Semi-Major and Semi-Minor Axes

In the given equation, a^2 is under x-term and b^2 is under y-term. Isolate these coefficients to find the square of the lengths of the semi-major axis (a) and semi-minor axis (b). Hence, a = radius = \(\sqrt{36/36} = 1\) (as \(36(x+4)^{2}\) equates to \(a^{2}\)) and b = \(\sqrt{36} = 6\) (as \((y+3)^{2}\) equates to \(36\)). notice that because the larger denominator corresponds to the y squared term, this means that the major axis is vertical.

Step 3: Calculate the Foci

The formula for foci when the major axis is vertical is \(c=\sqrt{b^2-a^2}\). Substituting the values of a and b, we get \(c=\sqrt{6^{2}-1^{2}} = \sqrt{35}\). The foci of the ellipse would therefore lie at \((-4, -3+\sqrt{35})\) and \((-4, -3-\sqrt{35})\). This is due to the ellipse being vertically aligned.

Step 4: Graph the Ellipse

Plot the center at \((-4,-3)\). Plot major axis from the center to \(6\) units up and down. Plot minor axis from the center to \(1\) unit left and right. Mark the foci at points \((-4, -3+\sqrt{35})\) and \((-4, -3-\sqrt{35})\). Then, connect these points in an oval shape to graph the ellipse.

Key Concepts

Ellipse Equation Standard Form

Understanding the standard form equation of an ellipse is fundamental for graphing and analysis. The general form follows the equation \(\frac{{(x-h)^{2}}}{{a^{2}}} + \frac{{(y-k)^{2}}}{{b^{2}}} = 1\), where \(h\) and \(k\) represent the coordinates of the ellipse's center.

For the given exercise, the standard form is somewhat obscured by the equation \(36(x+4)^{2}+(y+3)^{2}=36\). To make it resemble the standard form, one would typically divide each term by 36, simplifying it to \((x+4)^{2} + \frac{{(y+3)^{2}}}{{36}} = 1\). From this adjusted equation, it becomes apparent that the center of the ellipse is at \( (-4, -3) \), obtained by analyzing the \(h\) and \(k\) values in the context of \(x\) and \(y\) shifts from the origin.

Semi-Major and Semi-Minor Axes

In ellipse equations, \(a\) and \(b\) represent the lengths of the semi-major and semi-minor axes, respectively. These measures are pivotal as they define the ellipse's shape and size.

The semi-major axis is the longest radius of the ellipse, extending from the center to the furthest point on the curve. The semi-minor axis is the shortest radius, reaching from the center to the closest edge. If \(a > b\), the ellipse is stretched along the x-axis; conversely, if \(b > a\), as in the given exercise, the ellipse is stretched along the y-axis, making the major axis vertical. Here, \(a = 1\) and \(b = 6\), indicating a taller, vertically aligned ellipse rather than a wider, horizontally aligned one.

Ellipse Foci Calculation

To locate an ellipse's foci, one must understand that they are points along the major axis, equidistant from the center, where the sum of the distances from any point on the ellipse to the foci is constant. This property is crucial for many applications of ellipses, including in the paths of celestial bodies in astronomy.

The foci can be found using the equation \(c=\sqrt{b^2-a^2}\). In our exercise, substituting \(a = 1\) and \(b = 6\) into this formula yields \(c=\sqrt{35}\), which is the focal length from the center. Thus, for a vertically aligned ellipse, the foci will be at the coordinates \( (-4, -3\pm\sqrt{35}) \) because this is a vertical ellipse. These foci effectively dictate the curvature of the ellipse as they draw nearer to or further from each other.