Math Problem Statement

What is the standard equation of the ellipse with Foci at (-1,2) and (-1,4) and co-vertices at (-1 +- 2\sqrt{3})?

Solution

To find the standard equation of an ellipse, we need to determine its center, the lengths of its semi-major axis aa, and semi-minor axis bb, as well as its orientation (horizontal or vertical).

Step 1: Identify the Center of the Ellipse

The foci of the ellipse are given as (1,2)(-1, 2) and (1,4)(-1, 4). The center of an ellipse lies at the midpoint between the foci.

Center=(1+(1)2,2+42)=(1,3)\text{Center} = \left( \frac{-1 + (-1)}{2}, \frac{2 + 4}{2} \right) = (-1, 3)

Step 2: Determine the Distance Between the Foci (c)

The distance between the foci is 2c2c. The distance between the foci (1,2)(-1, 2) and (1,4)(-1, 4) is:

Distance=((1)(1))2+(42)2=0+22=2\text{Distance} = \sqrt{((-1) - (-1))^2 + (4 - 2)^2} = \sqrt{0 + 2^2} = 2

So, 2c=22c = 2, which gives c=1c = 1.

Step 3: Find the Length of the Semi-Minor Axis (b)

The co-vertices are given as (123,3)(-1 - 2\sqrt{3}, 3) and (1+23,3)(-1 + 2\sqrt{3}, 3). The distance between the co-vertices is 2b2b. The distance between the co-vertices is:

Distance=((1+23)(123))2=(43)2=43\text{Distance} = \sqrt{((-1 + 2\sqrt{3}) - (-1 - 2\sqrt{3}))^2} = \sqrt{(4\sqrt{3})^2} = 4\sqrt{3}

Therefore, 2b=432b = 4\sqrt{3}, giving b=23b = 2\sqrt{3}.

Step 4: Calculate the Length of the Semi-Major Axis (a)

We know that for an ellipse:

a2=b2+c2a^2 = b^2 + c^2

Substituting the values we found for bb and cc:

a2=(23)2+12=43+1=12+1=13a^2 = (2\sqrt{3})^2 + 1^2 = 4 \cdot 3 + 1 = 12 + 1 = 13

Thus, a=13a = \sqrt{13}.

Step 5: Write the Standard Equation of the Ellipse

Since the foci have the same xx-coordinate (-1), the ellipse is vertically oriented. The standard form for a vertical ellipse with center (h,k)(h, k) is:

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

Substituting the center (1,3)(-1, 3), a2=13a^2 = 13, and b2=12b^2 = 12:

(x+1)212+(y3)213=1\frac{(x + 1)^2}{12} + \frac{(y - 3)^2}{13} = 1

Final Answer

The standard equation of the ellipse is:

(x+1)212+(y3)213=1\boxed{\frac{(x + 1)^2}{12} + \frac{(y - 3)^2}{13} = 1}

Would you like a breakdown of each step or have any questions about the solution?

Related Questions:

  1. How do you find the foci of an ellipse given its equation?
  2. What is the difference between the standard forms of a horizontal and vertical ellipse?
  3. How do you derive the formula for the semi-major and semi-minor axes of an ellipse?
  4. What is the relationship between the eccentricity and the foci of an ellipse?
  5. How do you convert the equation of an ellipse from standard form to general form?

Tip: Remember that the distance between the foci and the length of the major axis are key factors in determining the shape and orientation of an ellipse.