ellipse problem help?

Question

Can anybody help with a quick ellipse problem?

Here's the info I have -- I'm told to find the equation of the ellipse:

Center at origin, passing through (2,2) and (3,-1), and having a coordinate axis as its principal axis.

Thanks!

Thanks! What a great help!

Answer 1

With center at the origin and the principal axes at the x-axis and y-axis, then the equation for the ellipse is:
x²/m² + y²/n² = 1

This is the general form for the ellipse. Since (x,y) it passes through (2,2) and (3,-1), then you can substitute them to the equation.
(2)²/m² + (2)²/n² = 1
(3)²/m² + (-1)²/n² = 1

Therefore,
4/m² + 4/n² = 1
9/m² + 1/n² = 1

We multiply by m²n²
4m² + 4n² = m²n²
m² + 9n² = m²n²

Solve for n² on both equations
m²n² - 9n² = m²
m²n² - 4n² = 4m²

Factor
n²(m² - 9) = m²
n²(m² - 4) = 4m²

Thus,
n² = m²/(m² - 9)
n² = 4m²/(m² - 4)

We then equate the two (TPE):
m²/(m² - 9) = 4m²/(m² - 4)

Cross-multiply
m²(m² - 4) = 4m²(m² - 9)

Transpose
m²(m² - 4) - 4m²(m² - 9) = 0

Factor out m²
m²[(m² - 4) - 4(m² - 9)] = 0

Since, from the original equation, m ≠ 0 and n ≠ 0, then we can divide m² from both sides
(m² - 4) - 4(m² - 9) = 0

We distribute
m² - 4 - 4m² + 36 = 0

Simplify
-3m² + 32 = 0

Then
3m² - 32 = 0

Then
m² = 32/3

Now, we have the value for m². To solve for n²,
n² = m²/(m² - 9)
n² = (32/3)/(32/3 - 9)
n² = (32/3)/(32/3 - 27/3)
n² = (32/3)/(5/3)
n² = 32/5

Therefore, the equation is:
x²/(32/3) + y²/(32/5) = 1

Or,
3x²/32 + 5y²/32 = 1

Or,
3x² + 5y² - 32 = 0

^_^

Answer 2

Kevin had a great and correct answer. The only thing I would co differently is the number cruching, which can be simplified.

With center at the origin and the principal axes at the x-axis and y-axis, then the equation for the ellipse is:
x²/m² + y²/n² = 1

This is the general form for the ellipse. Since (x,y) it passes through (2,2) and (3,-1), then you can substitute them to the equation.
(2)²/m² + (2)²/n² = 1
(3)²/m² + (-1)²/n² = 1

Therefore,
4/m² + 4/n² = 1
9/m² + 1/n² = 1

We multiply by m²n²
4m² + 4n² = m²n²
m² + 9n² = m²n²

Since both equations = m²n², set them equal.
4m² + 4n² = m² + 9n²
3m² = 5n²
m² = (5/3)n²

Since m and n are always going to be used when they are squared, we don't need to take the square root.

Plugging the first point into the equation we have:
4/[(5/3)n²] + 4/n² = 1

Multiplying by n²
4/[(5/3)] + 4 = n²

Grouping terms
4[ 3/5 +1] = n²
4(8/5) = n²
n² = 32/5

m² = (5/3)n²
m² = (5/3)(32/5) = 32/3

Therefore, the equation of the ellipse is:

x²/(32/3) + y²/(32/5) = 1

Or,
3x²/32 + 5y²/32 = 1

Or,
3x² + 5y² = 32

Answer 3

I'll bet the calculations can be even shorter than the last two. You know the form of an ellipse with the center at the origin and the axes aligned is:

(x/a)^2 + (y/b)^2 = 1

To make the equations linear, define: c = 1/a^2, d = 1/b^2
Now the form is: x^2c + y^2d = 1

Put in your two points and you get two linear equations:

4c + 4d = 1
9c + d = 1

Now subtract 4 times the second from the first:

(4c + 4d) - 4(9c + d) = 1 - 4
Simplifying: -32c = -3
c = 3/32

Solving for d: 9c + d = 1 so d = 1 - 9c = 1 - 27/32 = 5/32

Putting them back into the standard form: (3x^2)/32 + (5y^2)/32 = 1
Clearing out fractions: 3x^2 + 5y^2 = 32

Answer 4

7.5