Math Problem Statement

Change it to standard form and find the value of co-vertices, center, vertices, foci, and endpoints of LR. Draw the graph of the given ellipse equation: 16x^2 + 4y^2 - 32 = 0.

Solution

Let's break down the given equation and solve the problem step by step.

Given equation:

16x2+4y232=016x^2 + 4y^2 - 32 = 0

Step 1: Convert to standard form

First, add 32 to both sides: 16x2+4y2=3216x^2 + 4y^2 = 32

Now, divide the entire equation by 32 to make it easier to identify the form of the ellipse: 16x232+4y232=1\frac{16x^2}{32} + \frac{4y^2}{32} = 1 x22+y28=1\frac{x^2}{2} + \frac{y^2}{8} = 1

This is the standard form of the equation of an ellipse: x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 where a2=2a^2 = 2 and b2=8b^2 = 8, meaning a=2a = \sqrt{2} and b=22b = 2\sqrt{2}.

Step 2: Identify key features of the ellipse

  1. Center: Since there are no terms involving xx or yy shifts, the center of the ellipse is at (0,0)(0, 0).

  2. Vertices:

    • The vertices are located along the axis corresponding to the larger value of b2b^2, which is along the yy-axis.
    • The length of the semi-major axis (along yy) is b=222.83b = 2\sqrt{2} \approx 2.83.
    • Therefore, the vertices are at (0,±22)(0, \pm 2\sqrt{2}) or approximately (0,±2.83)(0, \pm 2.83).
  3. Co-vertices:

    • The co-vertices are along the xx-axis. The length of the semi-minor axis is a=21.41a = \sqrt{2} \approx 1.41.
    • Thus, the co-vertices are at (±2,0)(\pm \sqrt{2}, 0) or approximately (±1.41,0)(\pm 1.41, 0).
  4. Foci:

    • To find the foci, we need to calculate the value of cc, where c2=b2a2c^2 = b^2 - a^2. c2=82=6soc=62.45c^2 = 8 - 2 = 6 \quad \text{so} \quad c = \sqrt{6} \approx 2.45
    • The foci are located along the yy-axis at (0,±6)(0, \pm \sqrt{6}) or approximately (0,±2.45)(0, \pm 2.45).
  5. Endpoints of the Latus Rectum:

    • The endpoints of the latus rectum are located at a distance of b2a\frac{b^2}{a} from the foci along the xx-axis. 82=425.66\frac{8}{\sqrt{2}} = 4\sqrt{2} \approx 5.66
    • The endpoints of the latus rectum are at (±5.66,±2.45)(\pm 5.66, \pm 2.45).

Step 3: Graph of the Ellipse

You can sketch the ellipse with the following points:

  • Center at (0,0)(0, 0)
  • Vertices at (0,±2.83)(0, \pm 2.83)
  • Co-vertices at (±1.41,0)(\pm 1.41, 0)
  • Foci at (0,±2.45)(0, \pm 2.45)

Would you like further details on any specific part of the solution, or help with visualizing the graph?

Here are some related questions to explore:

  1. How is the general form of an ellipse derived?
  2. What is the significance of the semi-major and semi-minor axes?
  3. How do you calculate the eccentricity of an ellipse?
  4. What is the role of the latus rectum in conic sections?
  5. How can you generalize the approach to find the equation of an ellipse with a shifted center?

Tip: Always check whether the larger denominator is under x2x^2 or y2y^2 to determine the orientation of the ellipse.

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Math Problem Analysis

Mathematical Concepts

Ellipses
Conic Sections
Standard Form of an Ellipse

Formulas

Standard form of an ellipse: (x^2/a^2) + (y^2/b^2) = 1
c^2 = b^2 - a^2 (formula for foci)
Length of the latus rectum: b^2/a

Theorems

Equation of an ellipse
Pythagorean relation for ellipses

Suitable Grade Level

Grade 10-12