Vertex, focus, directrix question (1 Viewer)
- Thread starter BlueGas
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You have 4 standard positions for your parabolae, viz:
1) (x - h)2 = 4a(y - k) (upright parabola)
2) (x - h)2 = -4a (y - k) ('downright' parabola)
3) (y - k) 2 = 4a(x - h) (sideways parabola - open to the right)
4) (y - k)2 = -4a(x-h) (sideways parabola - open to the left)
Your parabola s of type-3. It is helpful to rewrite it in the standard form of type-3:
(y - 0)2 = 4 x 2 x (x - [-2])
So your 3 parameters are: h = -2, k = 0, a = 2
i) So your vertex (h,k) is simply (-2,0)
ii) Your focal length a = 2; so your Focus is 2 units from the vertex along the line that divides the parabola into 2 equal halves (the axis of symmetry, which in this case is simply the x-axis) and inside the parabola.
So the focus is simply (0,0), the origin.
iii)Your directrix is the straight line perpendicular to the axis of symmetry (so it is parallel to the y-axis), that is 2 units away from the vertex, on the opposite side of the vertex from the focus.
so your directrix is simply the equation: x = -4
iv) you should be able to sketch the parabola now
1) (x - h)2 = 4a(y - k) (upright parabola)
2) (x - h)2 = -4a (y - k) ('downright' parabola)
3) (y - k) 2 = 4a(x - h) (sideways parabola - open to the right)
4) (y - k)2 = -4a(x-h) (sideways parabola - open to the left)
Your parabola s of type-3. It is helpful to rewrite it in the standard form of type-3:
(y - 0)2 = 4 x 2 x (x - [-2])
So your 3 parameters are: h = -2, k = 0, a = 2
i) So your vertex (h,k) is simply (-2,0)
ii) Your focal length a = 2; so your Focus is 2 units from the vertex along the line that divides the parabola into 2 equal halves (the axis of symmetry, which in this case is simply the x-axis) and inside the parabola.
So the focus is simply (0,0), the origin.
iii)Your directrix is the straight line perpendicular to the axis of symmetry (so it is parallel to the y-axis), that is 2 units away from the vertex, on the opposite side of the vertex from the focus.
so your directrix is simply the equation: x = -4
iv) you should be able to sketch the parabola now
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davidgoes4wce
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Here is my answer:
Flop21
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Drongoski pretty much explained it. You gotta make it into the standard forms.
Quick question, are there any other types of forms I need to know for the parabola to find the vertex etc.???
davidgoes4wce
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@ Blue gas , think of the equation by finding the 'a' value first by breaking it up into separate parts:
y^2=8(x+2)
y^2= 4 x 2 x (x+2)
a=2 . The graph is being graphed horizontally, and with an 'a' value of 2, the directrix is 2 units 'outside' the parabola or 'away' from the vertex and the Focus is 2 units 'inside' the parabola.
Hope that clears it up.
y^2=8(x+2)
y^2= 4 x 2 x (x+2)
a=2 . The graph is being graphed horizontally, and with an 'a' value of 2, the directrix is 2 units 'outside' the parabola or 'away' from the vertex and the Focus is 2 units 'inside' the parabola.
Hope that clears it up.
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keepLooking
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Okay, let me try to explain it as easy as I can.
For example, you get an equation like this:
BlueGas
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Okay I'm starting to get this now, but howcome you done 2 + 1 = 3 to find the focus? In my original post when the focal length of 2 was found, the focus was the y coordinate of the vertex minus the focal length of 2, but in your example you done something different?Okay, let me try to explain it as easy as I can.
For example, you get an equation like this:
^2 = 4a(y-k)\\\\$Let's actually sub in numbers to make it easier.$\\(x-4)^2=4(y-2)\\$You know the vertex is V(4,2)$\\\\$You know that the equation I have given you is a concave up (by inspection). This will mean that the Directrix will have a lower y-coordinate than the Vertex. As for the Focus, it will have a greater y-coordinate than the vertex.$\\\\$You also know that the focal length is 1. Since $ a = 1. $ Since you know the Focus is $ a $ units away from the Vertex, you add $ a $ to the y-coordinate as the curve is concave up.$\\$The Focus becomes$ F(4,3) $(since a = 1, and $2+1=3$).$\\\\$You also know that the Directrix is a units away from the Vertex (in this case you subtract$ a $ from the Vertex's y-coordinate.$\\$The Directrix becomes y=1 ($2-1=1$).$)
BlueGas
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I'm getting a much better idea now, once I know the shape of the parabola I can find out the rest, but is there an easy way to know without knowing these equations?
1) (x - h)2 = 4a(y - k) (upright parabola)
2) (x - h)2 = -4a (y - k) ('downright' parabola)
3) (y - k) 2 = 4a(x - h) (sideways parabola - open to the right)
4) (y - k)2 = -4a(x-h) (sideways parabola - open to the left)
1) (x - h)2 = 4a(y - k) (upright parabola)
2) (x - h)2 = -4a (y - k) ('downright' parabola)
3) (y - k) 2 = 4a(x - h) (sideways parabola - open to the right)
4) (y - k)2 = -4a(x-h) (sideways parabola - open to the left)
I don't remember the equations fully, I just think it through. For an parabola that goes up, the y-values are increasing so 4a is positive and vice-versa for negative, and for the parabolas going to the right, the x-values are increasing so it's 4a is positive and vice versa for a parabolas going to the left.
You have to remember the 4a bit, but not exactly the (y-k) and (x-h). I just know that it will be x minus the x-coordinate of the vertex and y minus the y-coordinate of the vertex. Like a circle, it's x minus the x-coordinate of the centre of the circle and y minus the y-coordinate of the centre.