the extension is still available and if necessary will automatically be given

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** Completing the square on the left-hand side we have

y^2+12y = (y^2 + 12 y + 6^2) - 6^2 or

(y^2 + 12 y + 36) - 36, which we put in the form

(y+6)^2 - 36.

The equation becomes

(y+6)^2 - 36 = - x + 1, which we rearrange to get

(y - (-6))^2 = -x + 37.

The form we need is

(y-k)^2 = 4 a ( x - h).

The right-hand side is -x + 37 = -1(x - 37) = 4 ( -1/4) ( x - h), so the equation is

(y-k)^2 = 4 (-1/4) ( x - 37).

The vertex is therefore at (-6, 37). The focus lies at displacement -1/4 = -.25 in the x direction from the vertex, at (36.75, -6). The directrix is at x = 37 - (-.25) = 37.25. **

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** The parabola has form (y-h) = 1 / (4 a) * ( x - k )^2 with vertex (h, k) and vertex-to-focus displacement a.

The vertex is (1,-1) so we have (y - -1) = 1 / (4 a) * (x - 1) or (y + 1) = 1 / (4 a) * (x - 1)^2.

Substituting (0,1) into the equation we have (1 + 1) = 1 / (4a) * ( 0 - 1 )^2, or 2 = 1 / (4a) so that a = 1/8; this makes 1 / (4 a) = 1/(4 * 1/8) = 1 / (1/2) = 2.

Thus (y + 1) = 2 ( x - 1)^2.

We can verify this by noting that a parabola with vertex at (1,-1) and y-intercept (0,1) is just the graph of y = x^2 stretched by factor 2 to give y = 2 x^2 then shifted 1 unit right and 1 unit down to give (y + 1) = 2 ( x - 1)^2. **

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** for this ellipse a^2 = b^2 - c^2 = 2^2 - 1^2 = 3, so the correct equation is

x^2 / 3 + y^2 / 4 = 1. **

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** 9 x^2 + y^2 - 18 x = 0. Factoring the 9 out of the x terms we have

y^2 + 9 ( x^2 - 2x) = 0. Completing the square in the parentheses we get

y^2 + 9 ( x^2 - 2 x + 1 - 1) = 0, or

y^2 + 9 ( (x-1)^2 - 1) = 0, or

y^2 + 9(x-1)^2 - 9 = 0 or

y^2 + 9(x-1)^2 = 9. Dividing both sides by 9 we have

y^2 / 9 + (x-1)^2 / 1 = 1, or

y^2 / 3^2 + (x-1)^2 / 1^2 = 1.

The semiaxis in the x direction is 1, the semiaxis in the y direction is 3, the center is (1,0).

So the vertices lie 1 unit right and left of the center, and 3 units above and below the center.

This puts the vertices at (2, 0) and (0, 0), and at (1, -3) and (1, 3).

The coordinates of any of the vertices can be plugged into 9x^2+y^2-18x=0 to obtain an identity. For example, (1, 3) gives us 9(-1)^2 + 3^2 - 18 (1) = 9 + 9 - 18 = 0. Plugging in the coordinates of all 4 vertices verifies our results. **

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the aphelion is the greatest distance from the Sun and the perihelion is the closest approach. The Sun is at one of the foci of the ellipse.

The periohelion and aphelion are along the same axis, with the foci also along this axis. The Sun is at one of the foci.

To see these it's a good idea to draw a sketch of an ellipse. Let the major axis be the x axis (it could as well be the y axis, or in fact any straight line; but we lose nothing by assuming the axis to be the x axis) with the center of the ellipse at the origin. The ellipse goes through the x axis at (-a, 0) and (a, 0).

The Sun is at one of the foci; let the Sun be at the point (-c, 0).

From the point (-a, 0) to (-c, 0) is the perihelion, and from (-c, 0) to (a, 0) is the aphelion. Thus the perihelion is a - c and the aphelion is a + c.

So a is halfway between the perihelion and the aphelion--i.e., a is the mean distance of the Earth from the Sun.

a - c is the thus 128.5 million miles, a is 142 million miles and c is the difference 13.5 million miles.

The aphelion is therefore a + c = 155.5 million miles.

b is the semi-minor axis, with b^2 = a^2 - c^2 = 142^2 - 13.5^2 = 19982, so that b = sqrt(19982) = 141.36, with distances measured in millions of miles.

So an equation of the ellipse could be

x^2 / a^2 + y^2 / b^2 = 1, i.e.,

x^2 / (142)^2 - y^2 / (141.36)^2 = 1.

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See my notes. I recommend that you self-critique a copy of this document, marking insertions with ****.