Solutions To Analytic Geometry Problems (3): The Parabola

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: 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:



The  directrix is parallel to the y-axis and 3 units to the left of the vertex so its equation is x = - 6

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)