Help with a calulus question?

Question

Find an equation for the conic that satisfies the given conditions; Parabola, vertex (1,0) Directix x=-5

Answer 1

I'm going to show you how to get the answer without any formulas.

The focus is (6,0) since the directrix is x=-5. Since every point on the parabola is equidistant from the focus and the directix, we can choose a random point (x,y) and try to solve for it. The distance from the directrix to the point is: x+5. The distance from the point to the focus is sqrt((x-1)^2 + y^2). Since the two quantities are equal,

x+5 =sqrt((x-1)^2+y^2)
(x+5)^2 = (x-1)^2 + y^2
Isolating x,
(x+5)^2 - (x-1)^2 = y^2
(x+5+x-1)(x+5-x+1) = y^2
(2x+4)(6) = y^2
12x+24 = y^2
12x = y^2-24
x = y^2/12 - 2
That's your equation

Answer 2

x-1=(1/24)y^2

or

x= (1/24)y^2+1

Why? Well, the distance from the vertex to the directrix is 6 units, so the distance from the vertex to the focus must be 6 also.
Since the directrix is vertical, the parabola must open to the right; it always opens away from the directrix.

When a parabola opens right or left, its eqn. is

x-h = (1/4c)(y-k)^2, where (h,k) is its vertex. and c is the distance from the vertex to the focus. So we plug in 1 for h, 0 for k, and 6 for c.

Answer 3

second answer by jenh is right

third answer by vishal made a mistake of mixing vertex and focus.
(1,0) is given as the vertex not the focus

Answer 4

Answerz: use your TI-83