1) Graph the parabola (x - 3/4)2 = -5/ 2 (y + 23/ 40) and defined by the properties given above.
Solution:
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.
Solution:
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.
Solution:
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 = x 2 + 4x
Solution:
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.
Solution:
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)
applies, so:
(y - 1) 2 = 12 (x + 3 )
For which 4p = 12 and p = 3, so we need a directrix equation 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.
Solution:
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
Hence:
y - 1 = 2 (x -1) or: y = 2x - 1
And this tangent line is shown below for the point P1 (x1, y1) = (1,1)
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