Download More Problems on Conic Sections

More Problems on Conic Sections
by CHED on June 18, 2017
lesson duration of 4 minutes
under Precalculus
generated on June 18, 2017 at 09:28 am
Tags: Analytic Geometry
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Generated: Jun 18,2017 05:28 PM
More Problems on Conic Sections
( 4 mins )
Written By: CHED on July 2, 2016
Subjects: Precalculus
Tags: Analytic Geometry
Resources
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Content Standard
The learners demonstrate an understanding of key concepts of conic sections and systems of nonlinear equations
Performance Standard
The learners shall be able to model situations appropriately and solve problems accurately using conic sections and
systems of nonlinear equations
Learning Competencies
The learners will recognize the equation and important characteristics of the different types of conic sections
The learners solves situational problems involving conic sections
Identifying the Conic Section by Inspection 1 mins
Inspecting the equation can lead us to the right conic section for its graph, and set us on the right step towards
analyzing it. We will also look at problems that use the properties of the different conic sections, allowing us to
synthesize what has been covered so far.
The equation of a circle may be written in standard form
Ax2 + Ay2 + Cx + Dy + E = 0,
that is, the coefficients of x2 and y2 are the same. However, it does not follow
that if the coefficients of x2 and y2 are the same, the graph is a circle.
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For a circle with equation (x – h)2 + (y – k)2 = r2 , we have r2 > 0. This is not the case for the standard equations of (A)
and (B).
In (A), because the sum of two squares can only be 0 if and only if each square is 0, it follows that x – 1 / 2 = 0 and y +
3 / 2 = 0. The graph " is thus the single point ( 1 / 2 ,–3 / 2 ) .
In (B), no real values of x and y can make the nonnegative left side equal to the negative right side. The graph is then
the empty set.
Let us recall the general form of the equations of the other conic sections. We may write the equations of conic
sections we discussed in the general form
Ax2 + By2 + Cx + Dy + E = 0.
Some terms may vanish, depending on the kind of conic section.
(1) Circle: both x2 and y2 appear, and their coefficients are the same
Ax2 + Ay2 + Cx + Dy + E = 0
Example: 18x2 + 18y2 – 24x + 48y – 5 = 0
Degenerate cases: a point, and the empty set
(2) Parabola: exactly one of x2 or y2 appears
Ax2 + Cx + Dy + E = 0 (D ? 0, opens upward or downward)
By2 + Cx + Dy + E = 0 (C ? 0, opens to the right or left)
Examples: 3x2 – 12x + 2y + 26 = 0 (opens downward)
– 2y2 + 3x + 12y – 15 = 0 (opens to the right)
(3) Ellipse: both x2 and y2 appear, and their coefficients A and B have the same sign and are unequal
Examples: 2x2 + 5y2 + 8x – 10y –7 = 0 (horizontal major axis)
4x2 + y2 – 16x – 6y + 21 = 0 (vertical major axis)
If A = B, we will classify the conic as a circle, instead of an ellipse.
Degenerate cases: a point, and the empty set
(4) Hyperbola: both x2 and y2 appear, and their coefficients A and B have different signs
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Examples: 5x2 – 3y22 – 20x – 18y – 22 = 0 (horizontal transverse axis)
– 4x2 + y2 + 24x + 4y – 36 = 0 (vertical transverse axis)
Degenerate case: two intersecting lines
The following examples will show the possible degenerate conic (a point, two intersecting lines, or the empty set) as
the graph of an equation following a similar pattern as the non-degenerate cases.
Seatwork/Homework 1 mins
The graphs of the following equations are (nondegenerate) conic sections. Identify the conic section.
(1) 5x2 – 3y2 + 10x – 12y = 22
Answer: hyperbola
(2) 2y2 – 5x – 12y = 17
Answer: parabola
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(3) 3x2 + 3y2 + 42x – 12y = –154
Answer: circle
(4) 3x2 + 6x + 4y = 18
Answer: parabola
(5) 7x2 + 3y2 – 14x + 12y = –14
Answer: ellipse
(6) –4x2 + 3y2 + 24x – 12y = 36
Answer: hyperbola
Problems Involving Different Conic Sections 0 mins
The following examples require us to use the properties of different conic sections at the same time.
Example 1.5.1. A circle has center at the focus of the parabola y2+16x+4y = 44, and is tangent to the directrix of this
parabola. Find its standard equation.
Solution. The standard equation of the parabola is (y + 2)2 = –16(x – 3). Its vertex is V (3,–2). Since 4c = 16 or c = 4, its
focus is F(–1,–2) and its directrix is x = 7. The circle has center at (–1,–2) and radius 8, which is the distance from F to
the directrix. Thus, the equation of the circle is
(x + 1)2 + (y + 2)2 = 64.
Example 1.5.2. The vertices and foci of 5x2 – 4y2 + 50x + 16y + 29 = 0 are, respectively, the foci and vertices of an
ellipse. Find the standard equation of this ellipse.
Solution. We first write the equation of the hyperbola in standard form:
(x + 5)2 / 16 – (y – 2)2 / 20 = 1.
For this hyperbola, using the notations ah, bh, and ch to refer to a, b, and c of the standard equation of the hyperbola,
respectively, we have ah = 4, bh = 2?5,
= 6, so we have the following points:
center: (–5, 2)
vertices: (–9, 2) and (–1, 2)
foci: (–11, 2) and (1, 2).
It means that, for the ellipse, we have these points:
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center: (–5, 2)
vertices: (–11, 2) and (1, 2)
foci: (–9, 2) and (–1, 2).
In this case, ce = 4 and ae = 6, so that
of the ellipse is
. The standard equation
(x + 5)2 / 36 + (y ? 2)2 / 20 = 1.
Seatwork/Homework 1 mins
1. Find the standard equation of all circles having center at a focus of 21x2 – 4y2 + 84x ? 24y = 36 and passing through
the farther vertex.
Answer: (x + 7)2 + (y + 3)2 = 49, (x ? 3)2 + (y + 3)2 = 49
2. Find the standard equation of the hyperbola one branch of which has focus and vertex that are the same as those of
x2 – 6x + 8y = 23, and whose conjugate axis is on the directrix of the same parabola.
Answer: (y – 6)2 / 4 – (x – 3)2 / 12 = 1
Exercises 1 mins
1. The graphs of the following equations are non-degenerate conic sections. Identify the conic section.
(a) 5x2 + 7y2 – 40x – 28y = –73
Answer: ellipse
(b) 5y2 + 2x – 30y = –49
Answer: parabola
(c) 3x2 – 3y2 + 12x – 12y = 5
Answer: hyperbola
(d) 3x2 + 3y2 + 12x + 12y = 4
Answer: circle
(e) 2x2 + 24x – 5y = –57
Answer: parabola
2. The graphs of the following equations are degenerate conic sections. What are the specific graphs?
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(a) x2 + 3y2 – 4x + 24y = –52
Answer: point: (2,–4)
(b) 9x2 – 4y2 + 18x – 16y = 7
Answer: lines: y + 2 = ±3 2 (x + 1)
(c) 3x2 + 5y2 – 6x – 20y = –25
Answer: empty set
3. An ellipse has equation 25x2 + 16y2 + 150x – 32y = 159. Find the standard equations of all parabolas whose vertex
is a focus of this ellipse and whose focus is a vertex of this ellipse.
Answer: (x + 3)2 = –8(y + 2), (x + 3)2 = 32(y + 2), (x + 3)2 = –32(y – 4), and (x + 3)2 = 8(y – 4)
Solution. The standard equation of the ellipse is
(x + 3)2 / 16 + (y – 1)2 / 25 = 1.
Its center is (–3, 1). Since a = 5 and b = 4, we get c = 3, so the vertices are P(–3,–4) and S(–3, 6), while its foci are
Q(–3,–2) and R(–3, 4). We then get four parabolas satisfying the conditions of the problem. The focal distance
indicated below is the distance from the vertex to the focus.
4. Find the standard equation of the hyperbola whose conjugate axis is on the directrix of the parabola y2 +12x+6y =
39, having the focus of the parabola as one of its foci, and the vertex of the parabola as one of its vertices.
Answer: (x ? 7)2 9 ? (y + 3)2 27 = 1
Solution.
Solution. The standard equation of the parabola is (y + 3)2 = –12(x – 4), so its vertex is V (4,–3), and it opens to the left.
With 4c = 12, or c = 3, its focus is F(1,–3), and its directrix is x = 7. The hyperbola has its center on x = 7, its conjugate
axis, and a vertex at (4,–3). Its center is then C(7,–3). The conjugate axis is vertical so the hyperbola is horizontal, with
constants ah = CV = 3 and ch = CF = 6, so b2h = c2h – a2h = 27. The standard equation of the required hyperbola is
(x – 7)2 / 9 – (y + 3)2 / 27 = 1.
5. Find the standard equation of the parabola opening to the left whose axis contains the major axis of the ellipse x2 +
4y2 – 10x – 24y + 45 = 0, whose focus is the center of the ellipse, and which passes through the covertices of this
ellipse.
Answer: (y –3)2 = –4(x – 6)
Solution. The standard form of the ellipse is
(x – 5)2 / 16 + (y – 3)2 / 4 = 1.
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Its center (5, 3) is the focus of the parabola. Since b = 2, its covertices are W1(5, 1) and W2(5, 5). The vertex of the
parabola, c units to the right of (5, 3), is (5 + c, 3). Its equation can be written as (y –3)2 = –4c(x–(5 + c)). Since (5, 5) is
a point on this parabola, we have (5–3)2 = –4c(5–(5+c)). Solving this equation for c > 0 yields c = 1. Therefore, the
standard equation of the required parabola is (y – 3)2 = –4(x – 6).
6. Find the standard equation of the ellipse whose major and minor axes are the transverse and conjugate axes (not
necessarily in that order) of the hyperbola 4x2 – 9y2 – 16x – 54y = 29.
Answer: (x – 2)2 / 9 + (y + 3)2 / 4 = 1
Solution.
Solution. The standard equation of the hyperbola is
(y + 3)2 / 4 – (x – 2)2 / 9 = 1,
with center (2,–3), and constants ah = 2 and bh = 3. Since its conjugate axis (which is horizontal and has length 2bh =
6) is longer than its transverse axis (length 2ah = 4), the ellipse is horizontal. Its major axis has length 2ae = 6 and its
minor axis has length 2be = 4, so ae = 3 and be = 2. The ellipse shares the same center as the hyperbola. Thus, the
standard equation of the required ellipse is
(x ? 2)2 / 9 + (y + 3)2 / 4 = 1.
7. If m ? –3, 2, find the value(s) of m so that the graph of
(2m – 4)x2 + (m + 3)y2 = (m + 3)(2m – 4)
is
(a) a circle,
(b) a horizontal ellipse,
(c) a vertical ellipse,
(d) a hyperbola (is it horizontal or vertical?), or
(e) the empty set.
Answer: (a) m = 7, (b) 2 < m < 7, (c) m > 7, (d) –3 < m < 2 (horizontal), (e) m < –3
Solution. It might be helpful to observe that the equation is equivalent to
x2 / m + 3 + y2 / 2m ? 4 = 1.
(a) The graph is a circle if m+3 = 2m–4 > 0 (positive, so the graph is not a point or the empty set). This happens if m =
7.
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(b) We require 0 < 2m – 4 < m+ 3. Thus, 2 < m < 7.
(c) We require 0 < m+ 3 < 2m – 4. Thus, m > 7. (d) We need m+3 and 2m–4 to have different signs. We consider two
cases.
i. If m + 3 < 0 < 2m – 4, then m < –3 AND m > 2, which cannot happen.
ii. If 2m–4 < 0 < m+3, then –3 < m < 2. In this case, the equation can be written, with positive denominators, as x2 / m
+ 3 – y2 / 4 ? 2m = 1. The hyperbola is horizontal.
(e) The remaining case is when m < –3. In this case, m + 3 < 0 and 2m–4 < 0. This makes x2/(m+3)+y2/(2m–4)
negative, never equal to 1. The graph is then the empty set.
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