# Analytic Geometry

• July 10th 2008, 02:42 PM
Dave53
Analytic Geometry
I am lost on these questions.

1. X2=-4y Find the vertex, focus and directrix of the parabola and sketch the graph.

2. Find the equation in standard form of the parabola with focus (-2, 4) and directrix x = 4

3. Find the center, vertices, foci and sketch the graph of the ellipse with the following equation. 25x2+12y2=300

Thanks for helping me keep my sanity!!

Dave
• July 10th 2008, 04:00 PM
masters
Quote:

Originally Posted by Dave53
I am lost on these questions.

1. X2=-4y Find the vertex, focus and directrix of the parabola and sketch the graph.

Dave

$x^2=-4y$

Put in vertex form: $y=a(x-h)^2+k$
Vertex will be at $(h, k)$

$y=-\frac{1}{4}x^2$

$y=-\frac{1}{4}(x-0)^2+0$

Vertex is at $(0, 0)$

The equation of the directrix is $y = k-\frac{1}{4a}$

$y=0-\frac{1}{4(-\frac{1}{4})}\Longrightarrow y=1$

Focus is given by $(h, k+\frac{1}{4a})$

Therefore, the focus is $(0, 0+\frac{1}{4(-\frac{1}{4})}\Longrightarrow (0, -1)$
• July 10th 2008, 04:12 PM
Dave53
Thanks for the help!

I don't know if I'll ever understand this stuff.(Doh)
• July 10th 2008, 04:36 PM
masters
Quote:

Originally Posted by Dave53
Thanks for the help!

I don't know if I'll ever understand this stuff.(Doh)

Here:An Introduction To Conic Sections

You can find a pretty extensive tutorial of the conic sections (definitions, examples, etc) to help you understand them better.

I have to go now, but if you don't receive answers to your other questions you posted here, try posting them individually.

Later.
• July 10th 2008, 05:13 PM
ticbol
Quote:

Originally Posted by Dave53
I am lost on these questions.

1. X2=-4y Find the vertex, focus and directrix of the parabola and sketch the graph.

2. Find the equation in standard form of the parabola with focus (-2, 4) and directrix x = 4

3. Find the center, vertices, foci and sketch the graph of the ellipse with the following equation. 25x2+12y2=300

Thanks for helping me keep my sanity!!

Dave

1. X2=-4y Find the vertex, focus and directrix of the parabola and sketch the graph.

x^2 = -4y
y = -(1/4)x^2 ......a basic "vertical" parabola, whose vertex is at (0,0).

One general form of this basic vertical parabola is y = [1 /(4p)]x^2
where p is the focal distance from the vertex.
Also, p is the distance of the vertex from the directrix along the axis of symmetry of the parabola.

So, 1 /(4p) = 1/4
4p = 4
p = 1

Because the coefficient of the x^2 is negative...(it is -1/4 here)..., it means the parabola opens downward.

The vertex being at (0,0), then the axis of symmetry is the x=0 vertical line, or the y-axis.

Hence,
>>>focus is 1 unit below the vertex, or it is at (0,-1).
>>>directrix is 1 unit above the vertex, or it is the horizontal line y = 1.

I don't know how to graph/draw/sketch in computers/calculators.

--------------------------------------------
2. Find the equation in standard form of the parabola with focus (-2, 4) and directrix x = 4

Along the axis of symmetry, the vertex is equidistant from the focus and the directrix.

Since the givens say that the focus is at (-2,4) and that the directrix x=4 is a vertical line, then the parabola is "horizontal"...the axis of symmetry and the directrix are perpendicular always.
So the parabola is in the form (x -h) = [1 /(4p)](y -k)^2 ----(i)
where
(h,k) is the vertex
p = focal distance from the vertex.

We find p.
Along the axis of symmetry, or along the horizontal line y = 4,
absolute distance from focus to dirextrix is 2 +4 = 6 units.
half of that is 3 units.
That is the distance from the vertex to the focus.
That is p. Or, p = 3 units. ------------------------**
Also, the vertex is at point (1,4) ------------**

Hence, the parabola is
(x -1) = [1 /(4*3)](y -4)^2
x -1 = (1 /12)(y -4)^2 ----------------(to be revised)

EDIT:
Since the directrix (x = 4) is to the right of the focus (-2,4), then the horizontal parabola opens to the left. That means the coefficient of the y^2 must be negative.

------------------------------------------------------------
3. Find the center, vertices, foci and sketch the graph of the ellipse with the following equation. 25x2+12y2=300

25x^2 +12y^2 = 300 ---------(i)
Divide both sides by 300 to make the RHS = 1,
(25/300)x^2 +(12/300)y^2 = 1
(1/12)x^2 +(1/25) y^2 = 1
(x^2)/12 +(y^2)/25 = 1
(x^2)/[(2sqrt(3))^2] +(y^2)/[5^2] = 1 ------(ii)

That is a "vertical" ellipse in that the major axis is vertical.
semi-major axis, a = 5 units.
semi-minor axis, b = 2sqrt(3) units.
So, focal distance from center, c = sqrt(a^2 -b^2) = sqrt(25 -12) = sqrt(13) units.

Hence, for the ellipse,
>>>center is at (0,0).
>>>vertices are at (0,5) and (0,-5).
>>>foci are at (0,sqrt(13)) and (0,-sqrt(13)).