Download Math 237. Calculus II HW Solutions for 10.1 Assigned: 3, 5, 8, 11, 12

Math 237. Calculus II
HW Solutions for 10.1
Assigned: 3, 5, 8, 11, 12, 13, 15, 16, 19, 23, 26, 27, 30, 33, 37, 38
Selected for Grading: 3, 8, 11, 19, 26
Solutions.
3. The given equation is x 2 = –12y. This parabola opens downwardly. And p = 3. So the focus is the point
(0, –3) and the directrix has equation y = 3. Here is the sketch.

y






       

x














5. From the given equation, y 2 = x, I get that p = 1/4 and that the parabola opens to the right. So its focus is
the point (1/4, 0) and its directrix has equation x = –1/4. Here is the requested sketch.
y



x











8. First I'll rearrange the given formula into the appropriate standard form.
3x 2 – 9y = 0
x 2 = 3y = 4(3/4) y
So this parabola opens upwardly, and p = 3/4. The focus is (0, 3/4) and the directrix's equation is y = –3/4.
Here is the requested sketch.

y



x











11. Given: the directrix has equation y – 2 = 0. For just a little additional clarity I'll rewrite that as y = 2. This
tells me means that p = 2 and the parabola opens downwardly. So the form of its equation is x 2 = –4py
and the actual formula is x 2 = –8y.
12. From the fact that the focus is (0, –1/9), I glean that p = 1/9 and that the parabola opens upwardly. So its
equation has the form x 2 = 4py and the actual equation is
.
13. The focus is (–4, 0), so p = 4 and the parabola opens to the left. So its equation is y 2 = –16x.
15. Since the axis of the parabola is the x-axis, then the form of its equation is either y 2 = 4px or y 2 = –4px.
Since it passes through the point (3, –1), then it must open to the right, not the left. So the form of its
equation is y 2 = 4px. Plugging in x = 3 and y = –1 gives an equation to solve for p.
(–1)2 = 4p(3)
12p = 1
p = 1/12.
Simplifying 4p to 1/3 gives the answer: y 2 = (1/3) x.
Here is a sketch.

y

x










(3, –1)

16. Since the parabola's axis is the x-axis, its vertex is the origin, and it passes through the point (–2, 4), then it
opens to the left and its equation has the form y 2 = –4px.
42 = –4p(–2) = 8p
p = 2.
The equation is
. And here is a sketch.
y



  




x





19. Given: y 2 = 16x and the point (1, –4).
Equation of the tangent: Point is (1, –4). Slope:
. So m = –2.
Equation:
.
Equation of the normal line: Same point, slope is
, so the equation is
the sketch.
y

x







. Here is
23. The equation is
and the point in question is
.
For the equation of the tangent, the slope is again obtained by evaluating that derivative at that point.
,
,
,
So the equation of the tangent line at that point is
line is
and the equation of the normal
. Here is the sketch.

y
x
    






26. The equation is
and the point to be handled is
,
,
.
,
The equation of the tangent line is
and the equation of the normal line is
. Here is the sketch.

y



x
    


  




27. The slope of the tangent in general . . .
,
,
. . . is equal to
. At the point in question, i.e., for some as yet unknown y-value, we have
,
And so we also have
So the point is
. Here is the sketch.
y

x





30. Given any parabola in the plane, and any focal chord c, lay down an axis system such that the vertex of the
parabola is at (0, 0) and the parabola opens upwardly. Then the equation of the parabola is of the form x 2 =
4py, its focus is the point (0, p), and its directrix has equation y = –p. It might be easier to use the
equivalent form of the equation for this parabola,
. Here's a sketch for reference.
y
c
(0, p)
x
l
Here's the program: Find the equations of the two tangent lines. Use those equations to find where the
tangents intersect. Show that this point is on the directrix.
To find the equations of the two tangent lines, I need two points and two slopes. Let (a, a 2/4p) be the
coördinates of the chord's right endpoint. The coördinates of the other endpoint of the focal chord can be
found, as follows. We know two points on that chord: (a, a2/4p), and the focus, (0, p). The slope of the
focal chord is then
. And so the equation of the focal line is
find where this intersects the original parabola, replace y with
two solutions, one of them being x = a.) Here is the algebra.
The y-coördinate to go along with that second x-coördinate is
. To
and solve for x. (We should get
.
The equation of the tangent at the right endpoint is . . .
The equation of the tangent at the left endpoint is . . .
The point of intersection:
Solve the first equation for x:
Substituting this for x in the second equation:
Since the y-coördinate of the point of intersection is y = –p, and this is the equation of the directrix, then
this point of intersection is on the directrix.
33. To show that the vertex is the point on the parabola that is closest to the focus, I will express the distance
from any point on the parabola to the focus as a function of one variable and then use my Calc-I skills to
minimize that function. To start, let's assume that the formula for the parabola is the easier-to-use
or, even more simply,
. The distance from any point
to the focus
is
given by
To optimize this function on the interval (–∞, ∞):
, which is never undefined but is equal to zero when x = 0.
The first-derivative test:
For x < 0, d'(x) = x/p is negative. So the function d(x) is decreasing on (–∞, 0).
For x > 0, d'(x) = x/p is positive. So the function d(x) is increasing on (0, ∞).
Since d is decreasing on (–∞, 0) and increasing on (0, ∞), then it has its global minimum at x = 0.
So here's what we have. The distance from a point on the parabola to the focus is minimized at x = 0. The
corresponding point is P = (0, 0), which is the vertex.
37. Here is a sketch to match the drawing in the text. The parabola's equation is of the form y2 = 4px.
R
G
F
The coördinates of the point R are
for some 0 < x < 1. The coördinates of G are
. And the coördinates of F are (p, 0). So the distance in question is equal to
38. Here is another sketch. The parabola's equation is some unspecified y 2 = 4px.
P
A
The point P has coördinates
B
for some a > 0. So the coördinates of A are (a, 0). As for B . . .
Slope of the tangent at P:
, so
, so the slope of the tangent line is
, which
simplifies to
.
Equation of the normal line at P:
Since the slope of the tangent is
, then the slope of the normal line is
And the line crosses through the point
So the equation of the normal line is
.
.
.
The point B is the point on the normal line whose y-coördinate is zero. Plugging that into the above
equation and solving for x:
So, finally, the distance from A to B is just |AB| = (a + 2p) – a = 2p, which is a constant.