Ellipse: Center, Radius, Intersections & Graph

by TextBrain Team 47 views

Alright, let's dive into the world of ellipses! We've got a fun problem here: figuring out the center, radius (well, semi-major and semi-minor axes), and intersection points of an ellipse defined by the equation x2+4y2+8xβˆ’16yβˆ’4=0x^2 + 4y^2 + 8x - 16y - 4 = 0. Oh, and we'll graph it too! Buckle up, guys, it's gonna be a mathematical ride!

Finding the Center and Semi-Axes

The first step to unraveling this elliptical mystery is to rewrite the equation in standard form. This involves completing the square for both the xx and yy terms. Trust me, it's not as scary as it sounds!

So, our equation is x2+4y2+8xβˆ’16yβˆ’4=0x^2 + 4y^2 + 8x - 16y - 4 = 0. Let's group the xx and yy terms together:

(x2+8x)+(4y2βˆ’16y)=4(x^2 + 8x) + (4y^2 - 16y) = 4

Now, let's complete the square for the xx terms. To do this, we take half of the coefficient of the xx term (which is 8), square it (which is 16), and add it to both sides. For the yy terms, we first factor out the 4:

(x2+8x+16)+4(y2βˆ’4y)=4+16(x^2 + 8x + 16) + 4(y^2 - 4y) = 4 + 16

Next, complete the square for the yy terms inside the parenthesis. We take half of the coefficient of the yy term (which is -4), square it (which is 4), and add it inside the parenthesis. Since the parenthesis is multiplied by 4, we need to add 4 * 4 = 16 to the right side of the equation:

(x2+8x+16)+4(y2βˆ’4y+4)=4+16+16(x^2 + 8x + 16) + 4(y^2 - 4y + 4) = 4 + 16 + 16

Now we can rewrite the equation with squared terms:

(x+4)2+4(yβˆ’2)2=36(x + 4)^2 + 4(y - 2)^2 = 36

To get the standard form of an ellipse, we need to divide both sides by 36:

(x+4)236+(yβˆ’2)29=1\\\frac{(x + 4)^2}{36} + \\\frac{(y - 2)^2}{9} = 1

Alright! We've done it! Now we can easily identify the center and semi-axes.

The standard form of an ellipse centered at (h,k)(h, k) is:

(xβˆ’h)2a2+(yβˆ’k)2b2=1\\\frac{(x - h)^2}{a^2} + \\\frac{(y - k)^2}{b^2} = 1

Where:

  • (h,k)(h, k) is the center of the ellipse
  • aa is the semi-major axis (the longer radius)
  • bb is the semi-minor axis (the shorter radius)

Comparing our equation to the standard form, we can see that:

  • h=βˆ’4h = -4
  • k=2k = 2
  • a2=36a^2 = 36, so a=6a = 6
  • b2=9b^2 = 9, so b=3b = 3

Therefore, the center of the ellipse is (βˆ’4,2)(-4, 2), the semi-major axis is 6, and the semi-minor axis is 3. Awesome!

Finding the Intersection Points

Next up, let's find where this ellipse intersects the x and y axes. These are the points where either y=0y = 0 (x-axis intersection) or x=0x = 0 (y-axis intersection).

X-axis Intersections (y=0y = 0)

To find the x-axis intersections, we substitute y=0y = 0 into the standard form of our equation:

(x+4)236+(0βˆ’2)29=1\\\frac{(x + 4)^2}{36} + \\\frac{(0 - 2)^2}{9} = 1

Simplify:

(x+4)236+49=1\\\frac{(x + 4)^2}{36} + \\\frac{4}{9} = 1

Subtract frac49\\\\frac{4}{9} from both sides:

(x+4)236=1βˆ’49=59\\\frac{(x + 4)^2}{36} = 1 - \\\frac{4}{9} = \\\frac{5}{9}

Multiply both sides by 36:

(x+4)2=36βˆ—59=20(x + 4)^2 = 36 * \\\frac{5}{9} = 20

Take the square root of both sides:

x+4=Β±20=Β±25x + 4 = \\\pm \\\sqrt{20} = \\\pm 2\\\sqrt{5}

Solve for xx:

x=βˆ’4Β±25x = -4 \\\pm 2\\\sqrt{5}

So the x-axis intersection points are (βˆ’4+25,0)(-4 + 2\\\sqrt{5}, 0) and (βˆ’4βˆ’25,0)(-4 - 2\\\sqrt{5}, 0).

Y-axis Intersections (x=0x = 0)

Now, let's find the y-axis intersections by substituting x=0x = 0 into the standard form of our equation:

(0+4)236+(yβˆ’2)29=1\\\frac{(0 + 4)^2}{36} + \\\frac{(y - 2)^2}{9} = 1

Simplify:

1636+(yβˆ’2)29=1\\\frac{16}{36} + \\\frac{(y - 2)^2}{9} = 1

49+(yβˆ’2)29=1\\\frac{4}{9} + \\\frac{(y - 2)^2}{9} = 1

Subtract frac49\\\\frac{4}{9} from both sides:

(yβˆ’2)29=1βˆ’49=59\\\frac{(y - 2)^2}{9} = 1 - \\\frac{4}{9} = \\\frac{5}{9}

Multiply both sides by 9:

(yβˆ’2)2=5(y - 2)^2 = 5

Take the square root of both sides:

yβˆ’2=Β±5y - 2 = \\\pm \\\sqrt{5}

Solve for yy:

y=2Β±5y = 2 \\\pm \\\sqrt{5}

So the y-axis intersection points are (0,2+5)(0, 2 + \\\sqrt{5}) and (0,2βˆ’5)(0, 2 - \\\sqrt{5}).

Graphing the Ellipse

To graph the ellipse, we'll use the information we've gathered:

  • Center: (βˆ’4,2)(-4, 2)
  • Semi-major axis: a=6a = 6 (horizontal)
  • Semi-minor axis: b=3b = 3 (vertical)
  • X-axis intersections: (βˆ’4+25,0)(-4 + 2\\\sqrt{5}, 0) and (βˆ’4βˆ’25,0)(-4 - 2\\\sqrt{5}, 0)
  • Y-axis intersections: (0,2+5)(0, 2 + \\\sqrt{5}) and (0,2βˆ’5)(0, 2 - \\\sqrt{5})

Let's plot the center first. From the center, we move 6 units to the left and right to find the vertices along the major axis. We also move 3 units up and down from the center to find the co-vertices along the minor axis. Then we can sketch the ellipse by connecting these points in a smooth curve. The intersection points we calculated earlier help to refine the graph, ensuring it crosses the axes at the correct locations. Accurate plotting of these points will give you a precise visual representation of the ellipse.

Unfortunately, I can't draw a graph for you here. But you can use graphing software like Desmos or Geogebra, or even good old-fashioned graph paper, to plot these points and sketch the ellipse. Just enter the equation or plot the key points we found! Remember to label the center, vertices, co-vertices, and intersection points for a complete graph.

Conclusion

So, we successfully found the center, semi-axes, and intersection points of the ellipse defined by x2+4y2+8xβˆ’16yβˆ’4=0x^2 + 4y^2 + 8x - 16y - 4 = 0. We also discussed how to graph it. Great job, everyone! Understanding these steps allows you to analyze and visualize any ellipse given its equation. Keep practicing, and you'll become an ellipse expert in no time! Understanding the properties of ellipses is fundamental in various fields, including physics, astronomy, and engineering. Keep exploring!

I hope this detailed explanation helps! Let me know if you have any more questions.