Problem 45 - Precalculus with Limits

Question

In Exercises 29-52, identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. \(3x^2+y^2+18x-2y-8=0\)

Step-by-Step Solution

Answer

The given equation represents an ellipse with a center at (-3,1), vertices at (-3,4) and (-3,-2), foci at (-3, 1+sqrt(6)) and (-3, 1-sqrt(6)), and eccentricity of sqrt(6)/3.

Step 1: Identify the Conic Shape

From the given equation \(3x^2+y^2+18x-2y-8=0\), we can see that the coefficients of \(x^2\) and \(y^2\) are unequal, which means the equation represents an ellipse.

Step 2: Rewrite in Standard form

Rearrange the given equation in the form \(\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\) where \((h,k)\) are the coordinates of center. This can be achieved by completing the squares:\n\nGroup the \(x\)s and then the \(y\)s, move the constant to the right side: \(3(x^2+6x)+1(y^2-2y)=-8 \Rightarrow 3[(x+3)^{2}-9]+(y-1)^{2}=9\) => \(3(x+3)^{2}+(y-1)^{2}=9+9\) => \((x+3)^2/3 + (y-1)^2/9 =1 \).

Step 3: Identify the Center

Looking at the equation \((x+3)^2/3 + (y-1)^2/9 =1 \), the ellipse's center, \((h,k)\), is the point that makes both \(x-h\) and \(y-k\) equal to zero. From this equation, \(h=-3\) and \(k=1\), so the center is at point \((-3,1).\)

Step 4: Identify the Vertices

The vertices are situated a distance of \(a\) (semi-major axis) away from the center in both directions along the major axis. Since the denominator under \(y-1\)^2 is larger, the major axis is vertical. Starting from the center \((-3,1)\), move \(a=3\) units up and down to obtain the vertices. The vertices are \((-3,1+3)\) and \((-3,1-3)\) or points \((-3,4)\) and \((-3,-2)\)

Step 5: Calculate the Length of Semi-minor Axis

The semi-minor axis, represented as \(b\), is the square root of the denominator under \(x+3\)^2 in the standard form of equation. So, \(b= \sqrt{3}\)

Step 6: Calculate the Foci

The foci can be found using the relationship \(c=\sqrt{a^{2}-b^{2}}\), where \(c\) is the distance from either focus to the center. Substituting \(a=3\) and \(b= \sqrt{3}\), \(c=\sqrt{9-3}= \sqrt{6}\). The foci would be found on the major axis \(\sqrt{6}\) units away from the center. This gives coordinates \((-3, 1+\sqrt{6})\) and \((-3, 1-\sqrt{6})\)

Step 7: Find Eccentricity of the Ellipse

The eccentricity of an ellipse, denoted as \(e\), is given by the formula \(e=c/a = \sqrt{6}/3 \

Step 8: Sketch its graph

Plot the center, vertices, and foci on a graph and draw an ellipse through those points. The points \((-3, -2)\), \((-3, 4)\), \((-3, 1+\sqrt{6})\), and \((-3, 1-\sqrt{6})\) should all lie on the ellipse.