Solving conic sections

Question

An ellipse has an equation 25x² + 16y² + 150x - 32y = 159. Find the standard equations of all parabolas whose vertex is a focus of this ellipse and whose focus is a vertex of this ellipse. (There are four equations of parabola) parabola

Answer 1

We put the ellipse into standard form by first completing the square.

Thus,

25x² + 16y² + 150x - 32y = 159

25(x² + 6x) + 16(y² - 2y) = 159

25(x² + 6x + 9) - 25 * 9 + 16(y² - 2y + 1) - 16 * 1= 159

25(x² + 6x + 9) + 16(y² - 2y + 1) = 159 + 25 * 9 + 16 * 1

25(x² + 6x + 9) + 16(y² - 2y + 1) = 400

(x² + 6x + 9)/16 + (y² - 2y + 1)/25= 400

The center of our ellipse is at (-3, 1)

As √25 = 5, and this is with the y part of the equation, our vertices are

(-3, 1) ± (0, 5)

(-3, -4), (-3, 6)

Then, the focus is a distance√(25 - 16) = √9 = 3 from the center, also in the y direction, or

(-3, 1) ± (0, 3)

(-3, -2), (-3, 4)

We now have the vertices and centers of the 4 parabolas - vertex first

Note that the standard form of the parabola is (x - h)² = 4p(y - k).This is when the axis of the parabola is parallel to the y-axis, as it is in this problem. Otherwise, we switch the x and y in the equation. (you can switch the h and k, also, if you want the center to be at (h, k) )

(-3, -2); (-3, -4) p = -4 --2 = -2; Using the standard form, (x - -3)² = 4(-2)(y - -2) or (x + 3)² =-8(y + 2)

(-3, -2); (-3, 6) p = 6 --2 = 8; Using the standard form, (x - -3)² = 4(8)(y - -2) or (x + 3)² = 32(y + 2)

(-3, 4); (-3, -4) p = -4 - 4 = -8; Using the standard form, (x - -3)² = 4(-8)(y - 4) or (x + 3)² =-32(y - 4)

(-3, 4); (-3, 6) p = 6 - 4 = 2; Using the standard form, (x - -3)² = 4(2)(y - 4) or (x + 3)² = 8(y - 4)