Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) recognize the equation and important characteristics of the different types of conic sections; and
(2) solve situational problems involving conic sections.
Lesson Outline
(1) Conic sections with associated equations in general form (2) Problems involving characteristics of various conic sections (3) Solving situational problems involving conic sections Introduction
In this lesson, we will identify the conic section from a given equation. We will analyze the properties of the identified conic section. We will also look at problems that use the properties of the different conic sections. This will allow us to synthesize what has been covered so far.
1.5.1. Identifying the Conic Section by Inspection 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.
General Equation Standard Equation graph
(A) 2x2+ 2y2− 2x + 6y + 5 = 0 x −122
+ y +322
= 0 point (B) x2+ y2− 6x − 8y + 50 = 0 (x − 3)2+ (y − 4)2 = −25 empty set
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 − 12 = 0 and y + 32 = 0. The graph is thus the single point
1 2, −32.
DEPED COPY
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 6= 0, opens upward or downward) By2+ Cx + Dy + E = 0 (C 6= 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 dif-ferent signs
Examples: 5x2− 3y2− 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.
(1) 4x2+ 9y2− 16x + 18y + 25 = 0 =⇒ (x − 2)2
32 +(y + 1)2 22 = 0
=⇒ one point: (2, −1)
DEPED COPY
(2) 4x2+ 9y2− 16x + 18y + 61 = 0 =⇒ (x − 2)2
32 +(y + 1)2 22 = −1
=⇒ empty set (3) 4x2− 9y2− 16x − 18y + 7 = 0 =⇒ (x − 2)2
32 − (y + 1)2 22 = 0
=⇒ two lines: y + 1 = ±2
3(x − 2) A Note on Identifying a Conic Section
by Its General Equation
It is only after transforming a given general equation to standard form that we can identify its graph either as one of the degenerate conic sections (a point, two intersecting lines, or the empty set) or as one of the non-degenerate conic sections (circle, parabola, ellipse, or hyperbola).
1.5.2. Problems Involving Different Conic Sections
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. 2
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, ch =pa2h+ b2h = 6, so we have the following points:
DEPED COPY
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:
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 be = pa2e− c2e = √
20. The standard equation of the ellipse is
(x + 5)2
36 +(y − 2)2
20 = 1. 2
More Solved Examples
1. Identify the graph of each of the following equations.
(a) 4x2− 8x − 49y2+ 196y − 388 = 0 (b) x2+ 5x + y2− y + 7 = 0
(c) y2− 48x + 6y = −729
(d) 49x2 + 196x + 100y2 + 1400y + 196 = 0
(e) 36x2+360x+64y2−512y+1924 = 0
(f) x2+ y2− 18y − 19 = 0
(g) −5x2+ 60x + 7y2+ 84y + 72 = 0 (h) x2− 16x + 20y = 136
Solution:
(a) Since the coefficients of x2 and y2 have opposite signs, the graph is a hyperbola or a pair of intersecting lines. Completing the squares, we get
4x2− 8x − 49y2+ 196y − 388 = 0 4(x2− 2x) − 49(y2− 4y) = 388
4(x2− 2x + 1) − 49(y2− 4y + 4) = 388 + 4(1) − 49(4) (x − 1)2
49 − (y − 2)2 4 = 1.
Thus, the graph is a hyperbola.
(b) Since x2 and y2 have equal coefficients, the graph is a circle, a point, or the empty set. Completing the squares, we get
x2 + 5x + y2− y + 7 = 0
DEPED COPY
Since the right hand side is negative, the graph is the empty set.
(c) By inspection, the graph is a parabola.
(d) Since the coefficients of x2 and y2 are not equal but have the same sign, the graph is an ellipse, a point, or the empty set. Completing the squares, we get
Thus, the graph is an ellipse.
(e) Since the coefficients of x2 and y2 are not equal but have the same sign, the graph is an ellipse, a point, or the empty set. Completing the squares, we get
Since the right-hand side is 0, the graph is a single point (the point is (−5, 4)).
(f) Since x2 and y2 have equal coefficients, the graph is a circle, a point, or the empty set. Completing the squares, we get
x2+ y2− 18y − 19 = 0 x2+ y2− 18y + 81 = 19 + 81
x2+ (y − 9)2 = 100.
Thus, the graph is a circle.
(g) Since the coefficients of x2 and y2 have opposite signs, the graph is a hyperbola or a pair of intersecting lines. Completing the squares, we get
−5x2+ 60x + 7y2+ 84y + 72 = 0
DEPED COPY
−5(x2− 12x) + 7(y2+ 12y) = −72
−5(x2− 12x + 36) + 7(y2+ 12y + 36) = −72 − 5(36) + 7(36) (x − 6)2
7 − (y + 6)2 5 = 0.
Since the right hand side is 0, the graph is a pair of intersecting lines;
these are y + 6 = ±r 5
7(x − 6).
(h) By inspection, the graph is a parabola.
2. The center of a circle is the vertex of the parabola y2+ 24x − 12y + 132 = 0.
If the circle intersects the parabola’s directrix at a point where y = 11, find the equation of the circle.
Solution:
y2− 12y = −24x − 132 y2− 12y + 36 = −24x − 132 + 36
(y − 6)2 = −24x − 96 (y − 6)2 = −24(x + 4)
The vertex of the parabola is (−4, 6) and its directrix is x = 2. Thus, the circle has center (−4, 6) and contains the point (2, 11). Then its radius is p(−4 − 2)2 + (6 − 11)2 =√
61. Therefore, the equation of the circle is (x + 4)2+ (y − 6)2 = 61.
3. The vertices of the hyperbola with equation 9x2− 72x − 16y2− 128y − 256 = 0 are the foci of an ellipse that contains the point (8, −10). Find the standard equation of the ellipse.
Solution:
9x2− 72x − 16y2− 128y − 256 = 0 9(x2− 8x) − 16(y2+ 8y) = 256
9(x2− 8x + 16) − 16(y2+ 8y + 16) = 256 + 9(16) − 16(16) (x − 4)2
16 − (y + 4)2
9 = 1
The vertices of the hyperbola are (0, −4) and (8, −4). Since these are the foci of the ellipse, the ellipse is horizontal with center C(4, −4); also, the focal distance of the ellipse is c = 4. The sum of the distances of the point (8, −10) from the foci is
p(8 − 0)2+ (−10 − (−4))2+p
(8 − 8)2+ (−10 − (−4))2 = 16.
DEPED COPY
This sum is constant for any point on the ellipse; so 2a = 16 and a = 8. Then b2 = 82− 42 = 48. Therefore, the equation of the ellipse is
(x − 4)2
64 + (y + 4)2 48 = 1.
Supplementary Problems 1.5
For items 1 to 8, identify the graph of each of the following equations.
1. 9x2+ 72x − 64y2+ 128y + 80 = 0 2. 49x2− 490x + 36y2+ 504y + 1225 = 0 3. y2+ 56x − 18y + 417 = 0
4. x2+ 20x + y2− 20y + 200 = 0 5. x2− 10x − 48y + 265 = 0
6. −144x2− 1152x + 25y2− 150y − 5679 = 0 7. x2+ 4x + 16y2− 128y + 292 = 0
8. x2− 6x + y2 + 14y + 38 = 0
9. An ellipse has equation 100x2 − 1000x + 36y2 − 144y − 956 = 0. Find the standard equations of all circles whose center is a focus of the ellipse and which contains at least one of the ellipse’s vertices.
10. Find all parabolas whose focus is a focus of the hyperbola x2−2x−3y2−2 = 0 and whose directrix contains the top side of the hyperbola’s auxiliary rectangle.
11. Find the equation of the circle that contains all corners of the auxiliary rect-angle of the hyperbola −x2− 18x + y2+ 10y − 81 = 0.
12. Find the equations of all horizontal parabolas whose focus is the center of the ellipse 9x2+ 17y2− 170y + 272 = 0 and whose directrix is tangent to the same ellipse.
13. Find all values of r 6= 1 so that the graph of
(r − 1)x2+ 14(r − 1)x + (r − 1)y2− 6(r − 1)y = 60 − 57r is
(a) a circle,
DEPED COPY
(b) a point,
(c) the empty set.
14. Find all values of m 6= −7, 0 so that the graph of
2mx2− 16mx + my2+ 7y2 = 2m2− 18m is
(a) a circle.
(b) a horizontal ellipse.
(c) a vertical ellipse.
(d) a hyperbola.
(e) the empty set.
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