ellipse question?

Question

Find the equation of the ellipse that satisfy the given condition


heres the exact given taken from my quiz just 1 hour ago.

assuming that x^2/a^2 + y^2/a^2 is given. passing through (2,3), the latus rectum is 3 times the distance from the center to the focus.

how go get this?

assuming that x^2/a^2 + y^2/a^2 = 1 is given. passing through (2,3), the latus rectum is 3 times the distance from the center to the focus.

Answer 1

The question cannot be right as written. If x² and y² are both over a² then you have a circle and the focus is at the center. I assume you mean:

x²/a² + y²/b² = 1

The answer will be different if the major axis is vertical.

The circle passes thru (2,3). Plug in the values.

2²/a² + 3²/b² = 1
4/a² + 9/b² = 1
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Let
c = distance from center of ellipse to focus
2L = latus rectum

The latus rectum is 3 times the distance from the center to the focus.

2L = 3c
L = 3c/2

The distance from one focus to any point on the ellipse to the other focus is 2a.

If we choose the point on the ellipse vertically above one of the foci we have a right triangle. The hypotenuse is from the far focus to the point on the ellipse. The two legs are the distance between the two foci (2c) and the semi-latus rectum (L).

Use the Pythagorean Theorem.

(2a - L)² = L² + (2c)²

Substitute L = 3c/2.

(2a - 3c/2)² = (3c/2)² + (2c)²
4a² - 6ac + 9c²/4 = 9c²/4 + 4c²
4a² - 6ac = 4c²
4a² - 6ac - 4c² = 0
2a² - 3ac - 2c² = 0
(2a + c)(a - 2c) = 0
a = -c/2, 2c

Eliminate the negative solution since distances must be positive.

a = 2c
a² = 4c²

Solve for b².

b² = a² - c²
b² = 4c² - c²
b² = 3c²

b²/a² = 3c²/(4c²) = 3/4
b² = 3a²/4

Plug back into the original formula for the ellipse.

4/a² + 9/b² = 1
4/a² + 9/(3a²/4) = 1
4/a² + 12/a² = 1
4 + 12 = a²
a² = 16

b² = 3a²/4 = (3*16)/4 = 12

The equation of the ellipse is:

x²/a² + y²/b² = 1
x²/16 + y²/12 = 1

Answer 2

x^2/a^2 + y^2/b^2 = 1
Plug in (2,3),
2^2/a^2 + 3^3/b^2 = 1 ......(1)
Plug in (c, 3c/2), since the latus rectum is 3 times the distance from the center to the focus.
c^2/a^2 + (3c/2)^2/b^2 = 1......(2)
Add triangle relation,
a^2 = b^2+c^2......(3)

Solving the system of equations (1), (2), and (3) will give you the answer.
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x^2/a^2 + y^2/a^2 = 1 is not right because c ≠ 0.