Problem 52 - Thomas Calculus

Question

Exercises \(49-52\) give equations for ellipses and tell how many units up or down and to the right or left each ellipse is to be shifted. Find an equation for the new ellipse, and find the new foci, vertices, and center. $$ \frac{x^{2}}{16}+\frac{y^{2}}{25}=1, \quad \text { left } 4, \text { down } 5 $$

Step-by-Step Solution

Answer

The new equation is \( \frac{(x + 4)^2}{16} + \frac{(y + 5)^2}{25} = 1 \); center: \((-4,-5)\), vertices: \((-4,0)\), \((-4,-10)\); foci: \((-4,-2)\), \((-4,-8)\).

Step 1: Identify the Original Center, Vertices, and Foci

The given equation is \( \frac{x^2}{16} + \frac{y^2}{25} = 1 \). This is the equation of an ellipse centered at the origin \((0,0)\). The semi-major axis is along the y-axis with length 5, and the semi-minor axis is along the x-axis with length 4. Therefore, the vertices are \((0,\pm 5)\) and co-vertices at \((\pm 4, 0)\). The foci along the major axis use \(c = \sqrt{b^2 - a^2} = \sqrt{25 - 16} = 3\); hence, the foci locations are \((0, \pm 3)\).

Step 2: Calculate the New Center

According to the problem, the ellipse is shifted 4 units to the left and 5 units down. Shifting left 4 units moves the center from \((0,0)\) to \((-4,0)\). Shifting down 5 units further moves the center to \((-4, -5)\).

Step 3: Update the Equation of the Ellipse

The equation for an ellipse with a center \((h,k)\) is \( \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \), where \(a\) (semi-minor axis) is 4 and \(b\) (semi-major axis) is 5. Substituting \(h = -4\) and \(k = -5\), the new equation becomes \( \frac{(x + 4)^2}{16} + \frac{(y + 5)^2}{25} = 1 \).

Step 4: Update the Vertices

Vertices of the original ellipse were \((0, \pm 5)\). Shifting these vertices 4 units left and 5 units down: \((0 + -4, 5 + -5) = (-4,0)\) and \((0 + -4, -5 + -5) = (-4,-10)\). These are the new vertices \((-4,0)\) and \((-4,-10)\).

Step 5: Update the Foci

The original foci of \((0, \pm 3)\) are repositioned after shifting. Calculating \((0,3)\)'s reposition by shifting: \((0 - 4, 3 - 5) = (-4, -2)\). Same for the other focus: \((0 - 4, -3 - 5) = (-4, -8)\). The new foci are \((-4, -2)\) and \((-4, -8)\).

Key Concepts

Ellipse Equation

An ellipse equation in its standard form provides a useful way to describe the shape's geometry. The general equation of an ellipse centered at (\(h, k\)) is given by: \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] Here, \(a\) and \(b\) represent the lengths of the semi-minor and semi-major axes, respectively. If \(a < b\), the ellipse is vertically oriented, as seen in our original example where the equation \(\frac{x^2}{16} + \frac{y^2}{25} = 1\) characterizes a vertically stretched ellipse centered at (\(0,0\)). The horizontal semi-minor axis is 4 units, and the vertical semi-major axis is 5 units long.To find the transformed equation of an ellipse that's been shifted, replace \(h\) and \(k\) with the new center coordinates. For example, after a left shift by 4 and down shift by 5, the new center: (\(-4,-5\)) transforms the equation to: \[\frac{(x+4)^2}{16} + \frac{(y+5)^2}{25} = 1\] This equation now describes the relocated ellipse.

Foci

The foci of an ellipse are two special points located along the major axis, which help define the ellipse's shape and characteristics. They are positioned symmetrically around the center. The distance of each focus from the center, denoted by \(c\), can be calculated with the formula: \[c = \sqrt{b^2 - a^2}\] In our original problem, this gives \(c = \sqrt{25 - 16} = 3\). Therefore, the original foci are located at (\(0,3\)) and (\(0,-3\)).When you shift an ellipse, the foci shift similarly. By moving the original foci 4 units to the left and 5 units down, the new foci positions become (\(-4, -2\)) and (\(-4, -8\)).These new foci represent the same geometric significance relative to the new center of (\(-4, -5\)). It's essential to correctly identify and apply these shifts so that the ellipses retain their shape and properties.

Vertices

The vertices of an ellipse are the points where the major axis intersects the ellipse. Vertices determine the maximum extent of the ellipse in any direction. For the ellipse \(\frac{x^2}{16} + \frac{y^2}{25} = 1\), the vertices were originally at (\(0,5\)) and (\(0,-5\)), corresponding to the semi-major axis along the y-axis.
The step of shifting the ellipse involves recalculating the vertices, while considering the given horizontal and vertical shifts.* Shifting the vertex (\(0,5\)) by 4 units left and 5 units down results in (\(-4,0\)).* Similarly, shifting the vertex (\(0,-5\)) gives (\(-4,-10\)). Identifying new vertex positions allows for the complete description of the transformation in terms of size and placement of the ellipse.

Center Shifting

Shifting the center of an ellipse is a crucial operation affecting every aspect of the ellipse's description, from equation to element placement like foci and vertices. In dealing with shifts, a clear understanding of what happens to coordinates is important.Initially, the ellipse is centered at the origin (\(0,0\)). Shifting the ellipse left by 4 units alters the center to (\(-4,0\)); shifting it down by 5 units further adjusts the center to (\(-4,-5\)). Every element of the ellipse must be recalculated in light of this move. This impact is apparent on the equation, where transformed center coordinates become essential for accurate repositioning:* Translate center from (\(0,0\)) to (\(-4,-5\))* Apply these translations uniformly to all relevant parts, ensuring the ellipse maintains its structure. Ellipses react predictably to defined translations, preserving their shape while simply relocating to reflect new positional definitions.