__Problems__:

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: x

^{2}

^{}= 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: d

^{2}

^{}y / dx

^{2}

^{}= 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:

x

^{2}

^{}+ 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 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:

_{}

^{}

_{}

^{}

*3 units to the left of the vertex*so its equation is x = - 6

6) Find the tangent to the parabola y = x

^{2}

^{ }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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