Calculus: Conic Sections?

Question

For the following problem we are asked to find an equation for the conic that satisfies the given conditions:

1. Ellipse, foci (0, -1), (8, -1), vertex (9, -1)

For the solution I got:

[(x – 4)^2]/25 + [(y + 1)^2]/3 = 1

Is this correct?

Answer 1

You are almost right. You divided the second term by 3 instead of 3².

Center is the midpoint of the foci.
(h, k) = (4, -1)

c = distance center to focus
a = distance center to vertex

c = 8 - 4 = 4
a = 9 - 4 = 5
a² = 25

b² = a² - c² = 25 - 16 = 9

The line of symmetry for the ellipse goes thru the foci. Since the line of symmetry is horizontal, the major axis is horizontal.

The equation of the ellipse is:

(x - h)²/a² + (y - k)²/b² = 1

Plug in values.

(x - 4)²/25 + (y + 1)²/9 = 1

Answer 2

The midpoint of foci (0, -1) and (8, -1) = (4, -1) is the centre of the ellipse.
Shifting origin to (4, -1),
x = x' + 4, y = y' - 1
and vertex (9, -1) = (5, 0)
=> a = 5 and ae = 4 => e = 4/5
=> b = a √( 1 - e^2) = 5 √( 1 - 16/25) = 3
=> eqn. of ellipse w.r.t. to new coordinates is
x'^2 / 25 + y'^2 / 9 = 1
Replacing x' by x - 4 and y' by y + 1,
the required equation of the ellipse is
[(x - 4)^2 / 25 + (y + 1)^2 / 9] = 1.

[Note the small error in your answer. b^2 = 9 and not 3.]

Answer 3

your second denominator should have been 9