How to find a parabola when given focus and directrix?

Pirate Lady - 2007-06-29 09:50:12 - Mathematics

Find the equation of the parabola whose focus is at (3, 3) and whose directrix is at x = 7. This seems like it should be really easy but I'm having a hard time with it. Can someone help? Thank you!

Best Answer:

Hey there! The standard equation for the parabola, is shown below. y-k=1/4p*(x-h)^2 or x-h=1/4p*(y-k)^2 Since you mentioned the directrix, in the form of x=h-p, we have to use the second equation. The reason is showed below. y-k=1/4p*(x-h)^2, then vertex=(h,k), focus=(h,k+p), directrix=y=k-p x-h=1/4p*(y-k)^2, then vertex=(h,k), focus=(h+p,k), directrix=x=h-p. The focus and the directrix were given, but not the vertex. In that case, we have to use the midpoint formula. M=((x1+x2)/2),((y1+y2)/2) --> M=((3+7)/2),((3+3)/2) --> M=(10/2),(6/2) --> M=(5,3) So the vertex is (5,3). We need to find p. Use the formula for calculating the directrix. x=h-p --> 7=5-p --> 2=-p --> p=-2 Now substituting the values into the standard equation of a parabola, we get. x-h=1/4p*(y-k)^2 --> x-5=1/4(-2)*(y-3)^2 --> x-5=-1/8*(y-3)^2 So the standard equation of the parabola is x-5=-1/8*(y-3)^2. I hope it helps!

Answer:

There's an equation to figure this out. The vertex is halfway between the focus and the directrix. So the vertex must be (5,3). The equation is: x-5 = 1/(4p)(y-3)^2 where p is the distance from the directrix to the vertex. So for us p = 2, and we have: x-5 = (1/8)(y-3)^2

Since the focus is (3, 3) and a general point on the directrix can be given by (7, y), the distance formula would give us: (x - 3)^2 + (y - 3)^2 = (x - 7)^2 + (y - y)^2 Expanding this expression yields: x^2 - 6x + 9 + y^2 - 6y + 9 = x^2 - 14x + 49 Simplifying by subtracting the common terms from both sides gives: y^2 = -8x + 31 Solving this equation for x will produce this equation for our parabola: x = (y^2 - 31) / -8

The distance from the focus to any point on a parabola is equal to the distance from the point to the directrix. Take any point (x,y) The distance from the focus to the point is d^2 = (x-3)^2 + (y-3)^2 the distance from the point to the nearest point on the directrix is d^2 = (x-7)^2 equating these (x-3)^2 + (y-3)^2=(x-7)^2 (y-3)^2 = x^2 -14x + 49 -x^2+6x-9 (y-3)^2 = -8x+40 x = -1/8(y-3)^2+5 is the parabola Note that it is laying sideways, as you can tell from the line defining the directrix.

Hey there! The standard equation for the parabola, is shown below. y-k=1/4p*(x-h)^2 or x-h=1/4p*(y-k)^2 Since you mentioned the directrix, in the form of x=h-p, we have to use the second equation. The reason is showed below. y-k=1/4p*(x-h)^2, then vertex=(h,k), focus=(h,k+p), directrix=y=k-p x-h=1/4p*(y-k)^2, then vertex=(h,k), focus=(h+p,k), directrix=x=h-p. The focus and the directrix were given, but not the vertex. In that case, we have to use the midpoint formula. M=((x1+x2)/2),((y1+y2)/2) --> M=((3+7)/2),((3+3)/2) --> M=(10/2),(6/2) --> M=(5,3) So the vertex is (5,3). We need to find p. Use the formula for calculating the directrix. x=h-p --> 7=5-p --> 2=-p --> p=-2 Now substituting the values into the standard equation of a parabola, we get. x-h=1/4p*(y-k)^2 --> x-5=1/4(-2)*(y-3)^2 --> x-5=-1/8*(y-3)^2 So the standard equation of the parabola is x-5=-1/8*(y-3)^2. I hope it helps!