1) Graph the parabola (x - 3/4)2 = -5/ 2 (y + 23/ 40) and defined by the properties given above.
Close inspection of the graph will show the following properties satisfied:
a) The vertex is at V(3/4, - 23/40)
b) The axis of symmetry is x = h or x = 3/4..
c) The focus is p = 5/8 units below the vertex at F(3/4, - 6/5)
2)The parabola shown below:
Has the general form (y - k) 2 = - 4p (x - h)
Use this information to: a) find the actual equation of the curve, b) the coordinates of the focus, c) the equation for the directrix.
On inspection, we see the parabola opens to the right, hence the directrix equation must be: x = - p
Then the general form must actually be rewritten:
(y - k) 2 = - 4 (-p) (x - h) = (y - k) 2 = 4p (x - h)
We see the vertex of the parabola is at V(0, 0).
Thus: h = 0 and k = 0
Or, re-writing with the substitutions: y 2 = 4p x
Since 4p = 4 then p = 1 and the eqn. for directrix is: x = - 1
The focus F is at F(p, 0) or F(1, 0)
3) Find the focus and the directrix for the parabola: x2 = 8y
And graph it, clearly labeling the axis of symmetry.
Take the 2nd derivative before graphing and establish whether the parabola is oriented concave downward or upward.
Here: 4p = 8 so that p = 2
The slope of the tangent at any point is dy/dx = x/ 2p. which is 0 at the origin. But the second derivative: d2 y / dx2 = 1/2p which is positive, so the curve is concave upward. This is verified from the sketch of the graph below with the axis of symmetry (at y = 0) in bold:
The focus for this parabola is at F(0, -p) or F (0, -2)
The equation for the directrix is y = -2
4) Do the same for: y = x2 + 4x
Complete the square to obtain:
x2 + 4x + 4 = y + 4 = (x + 2) 2
But: (x - h) 2 = 4p (y - k)
So: h = -2 and k = -4
So: (x + 2) 2 = (y + 4)
The graph is shown below:
Here the axis of symmetry is at x = -2.
Since 4p = 1 and p = 1/4, the directrix is parallel to the x-axis and 1/4 unit below the vertex. (its equation is y = - 4 1/4). The focus F is on the axis of symmetry 1/4 unit above the vertex.
The coordinates are F(-2, - 3 3/4)
5) Given V(-3, 1) and F (0, 1) find the equation of the associated parabola and also for its directrix. Sketch the graph showing the focus, vertex and directrix.
From the vertex coordinates we have: h = -3 and k = 1
With these values this indicates a parabola opening to the right so the general form:
(y - k) 2 = 4p (x - h)
(y - 1) 2 = 12 (x + 3 )
For which 4p = 12 and p = 3, so we need a directrix eqn. of form x = -p
(But remember the x value is adjusted based on where the vertex is, so it is already at x = -3!)
The graph is shown below:
6) Find the tangent to the parabola y = x2 using the fact that the equation of the line tangent to a curve y = f(x) is given by:
y - y1 = f'(x1) (x - x1)
where (x1, y1) = P1 is the point of the curve for which the tangent is constructed and y1 = (x1)2 . Show the graph of the given parabola and the tangent line with the point P1 (x1, y1) identified.
f'(x1) = dy/ dx = 2x
So: y - y1 = 2x1 (x - x1)
But only one point satisfies y1 = (x1)2
That is x1 = 1, y1 = 1
y - 1 = 2 (x -1) or: y = 2x - 1
And this tangent line is shown below for the point P1 (x1, y1) = (1,1)