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Unformatted text preview: Conics Copyright © 2006, Barry Mabillard.
0 Conics Standards Test  ANSWERS www.math40s.com
www.math40s.com 1. Sketch the graph and write the equation of a circle with a centre at (3, 2) and is
tangent to the x – axis.
Start with
2
2
( x  h) + ( y  k ) = r 2
The phrase
“tangent to
the xaxis”
means the
circle touches
the xaxis. The centre is (3, 2), and
the radius is 2. Plug these
into the standard form
equation. ( x  3 ) + ( y  (2 ) ) = 22
2
2
( x  3 ) + ( y +2 ) = 4
2 2 2. Identify the conic represented by the equation x 2 − 6 x = y + 3
Since the y2 term is missing, the conic is a parabola.
3. State the coordinates of the vertex in the relation y 2 = 2 x + 4
y 2 = 2x + 4
y 2 = 2 ( x + 2) The vertex is located at (2, 0)
4. Change the following to standard form, then sketch: 4 y 2 + 40 y − 4 x 2 + 32 x + 20 = 0
4y 2 + 40y  4x 2 + 32x + 20 = 0
4 ( y 2 +10y ) 4( x
4 ( y +10y + 25 )  4 ( x 2 2 2 ) = 20
 8x +16 ) = 20+100  64  8x 4 ( y +5 )  4 ( x  4 ) = 16
2 2 4 ( y +5 ) 4 ( x  4 )
16
=
16
16
16
2 ( y +5 )
4 2  2 ( x  4)
4 2 =1 This is a vertical hyperbola since y comes first
Centre = (4, 5), a = 2, b = 2 Conics Standards Test  ANSWERS 1 www.math40s.com 5. a) Given the parabola below, find the equation if the vertex is at (4, 2)
The vertex is
located at (4, 2).
These are the
h & k values. Standard form for horizontal
parabolas → x  h= a ( y  k )
0  4 = a ( 0  (2 ) ) A point on the
graph is located
at (0, 0) This is a
value you can use
for x & y. 0  4 = a ( 0+ 2 ) 2 2 2 4= a ( 4 )
a= 1 Plug everything in
and solve for a. The equation is : x  4 =  ( y +2 ) b) Verify the intercepts algebraically x – intercept:
Set y = 0, then solve for x.
2
x  4 =  ( 0+ 2 ) y – intercepts:
Set x = 0, then solve for y. 0  4=  ( y + 2 ) 4=  ( y + 2 ) x  4 = 4
x =0 4= ( y + 2 )
4= 2 2 First evaluate for + 2
+2 = y + 2
y =0 2 (y + 2) ±2= y + 2 2 Then evaluate for  2
2 = y + 2
y = 4 6. A conic is represented by x 2 − 4 x + 9 y 2 − 18 y = 23
a) Sketch the conic
x 2  4x +9y 2  18y = 23 (x
(x 2
2 ) +9 ( y
 4x + 4 ) +9 ( y
 4x 2 2  2y ) = 23  2y +1) = 23+ 4+9 ( x  2 ) +9 ( y  1) = 36
2
2
( x  2 ) + 9 ( y  1) = 36
2 2 36 ( x  2)
36 36 2 + ( y  1)
4 36 2 Centre = (2, 1)
a = 6, b = 2 =1 b) State the domain & range
Domain is { x 4 ≤ x ≤ 8} Range is {y 1 ≤ y ≤ 3} Conics Standards Test  ANSWERS 2 www.math40s.com 2 7. Identify the conic with the equation 3 x 2 − y 2 − 7 x + 2 = 0
Since the product of A & C is negative, the conic is a hyperbola.
8. The equation of a circle is given by the equation 2 x 2 + 2 y 2 − 8 x + 4 y − 22 = 0
a) Sketch the conic
2x 2 + 2y 2  8x + 4y  22 = 0
2x 2  8x + 2y 2 + 4y = 22
2 ( x 2  4x ) + 2 ( y + 2y ) = 22
2 ( x  4x + 4 ) + 2 ( y + 2y +1) = 22+8+ 2
2 2 2 2 ( x  2 ) + 2 ( y +1) = 32
2 2 2 ( x  2)
2 ( y +1)
32
+
=
2
2
2
2 ( x  2) 2 2 + ( y +1) = 16
2 b) State the radius
2
2
The form is ( x  h ) + ( y  k ) = r 2 . Taking the square root of the right side gives 4 units. 9. Determine the equation of the ellipse shown below The a – value is 3, and the b – value is 2.
The centre is (0, 0)
Plug into the equation
2
2
( x  h) + ( y  k ) = 1
a2
b2 (x 0)
32 2 (y  0)
+
22 2 =1 x2 y2
+
=1
9
4 Conics Standards Test  ANSWERS 3 www.math40s.com 10. The equation of a conic is ( x − 2)
9 2 ( y + 2)
−
4 2 =1 a) Identify this conic section
The minus between the terms indicates this is a hyperbola
b) Sketch a clearly labeled graph Centre = (2, 2)
a = 3, b = 2 c) State the domain & range Domain is { x  x ≤ 1, x ≥ 5}
Range is {y y ∈ R} 11. Find the coordinates of the vertices and sketch the equation: 9x2 + 4y2 + 40y + 64 = 0 9x 2 +4y 2 +40y +64 = 0
9x 2 +4(y 2 +10y)=64
9x 2 +4(y 2 +10y + 25)=64+100
9x 2 4(y +5)2 36
=
+
36
36
36
2
2
x
(y +5)
+
=1
4
9 Conics Standards Test  ANSWERS 4 www.math40s.com 12. Find the equation of the ellipse sketched below: ( x  h) 2 (y  k )
+ 2 =1
a2
b2
Centre = (2, 1) ; a = 2, b = 3 Start with: ( x  2 ) + ( y +1) = 1
2
2
( 2)
(3)
2
2
( x  2 ) + ( y +1) = 1
2 2 4 13. Sketch the equation and state the domain and range for 9 ( x − 4) 2 4 − y2
=1
9 Centre = (4, 0) ; a = 2, b = 3 Domain is { x  x ≤ 2, x ≥ 6}
Range is {y y ∈ R} y 2 x2
−
=1
14. Sketch the equation and state the axis of symmetry:
16 25 This is a vertical hyperbola since y comes first.
Since the centre is at the origin, the xaxis
is the axis of symmetry. Conics Standards Test  ANSWERS 5 www.math40s.com 15. The equation of a conic is 25 x 2  9 y 2 100 x + 72 y  269 = 0
a) Write the equation in standard form
b) Sketch the graph
25x 2 9y 2  100x +72y  269 = 0
25x 2  100x 9y 2 +72y = 269
25 ( x 2  4x ( ) 9 ( y 2  8y )( ) ) = 269 2
2
25 x  4x + 4  9 y  8y + 16 = 269 + 100  144 25 ( x  2 ) 9 ( y  4 ) = 225
2 2 25 ( x  2 ) 9 ( y  4 )
225
=
225
225
225
2 ( x  2)
9 2  2 (y  4)
25 2 =1 Centre = (2, 4)
a = 3, b = 5 16. An ellipse has the following vertices: A(0, 9), B(2, 5), C(0, 1), and D(2, 5)
Draw the ellipse and determine the equation. The a – value is 2, and the b – value is 4.
The centre is (0, 5)
Plug into the equation
2
2
( x  h) + ( y  k ) = 1
a2
b2 ( x 0)
22 2 (y  5 )
+
42 2 =1 x 2 (y  5 )
+
=1
4
16
2 Conics Standards Test  ANSWERS 6 www.math40s.com 17. For the conic 2 y 2 − 2 x − 4 y − 6 = 0
a) Find the intercepts of the conic section y – intercepts:
Set x = 0, then solve for y.
2y 2  2 ( 0 )  4y  6 = 0 x – intercept:
Set y = 0, then solve for x.
2
2 ( 0 )  2x  4 ( 0 )  6 = 0 2y 2  4y  6 = 0
2 ( y 2  2y  3 ) = 0 2x = 6 y 2  2y  3 = 0 x = 3 ( y  3 )( y +1) = 0
y = 1, 3 b) Sketch a clearly labeled graph 2y 2  2x  4y  6 = 0
2y 2  4y = 2x +6
2 ( y 2  2y ) = 2x +6
2 ( y 2  2y +1) = 2x +6+ 2
2 ( y  1) = 2x +8
2 2 ( y  1) = 2 ( x + 4 )
2 ( y  1) 2 = ( x +4 ) 18. The conic x 2 + y 2 = 1 is translated 2 units to the right
a) Write the equation of the new conic and sketch it A translation 2 units right can be
accomplished by replacing x with x – 2. ( x  2) 2 +y 2 = 1 b) State the domain of the new conic
{ x 1 ≤ x ≤ 3}
c) State the range of the new conic
{y1 ≤ y ≤ 1} Conics Standards Test  ANSWERS 7 www.math40s.com ...
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 Spring '12
 A.Young
 Biology, pH

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