Name: DONOR, Mathew S. Instructor: Engr.

Marvin Radaza Practice Problems

Yr & Sec: BSEE 3 – H EE Comprehensive Course 1 - Mathematics Analytic Geometry

SET A (PART 1) PROBLEM #33: A line with inclination of 45°

5 9
PROBLEM #13: Find the angle between the passes through (− 2 , − 2). What is the x-
lines 3𝑥 + 2𝑦 = 6 and 𝑥 + 𝑦 = 6. coordinate of a point on the line if its
responding y-coordinate is 6?
A. 12°20′ C. 14°25′
B. 𝟏𝟏°𝟏𝟗′ D. 13°06′ A. 6 C. 𝟖
B. 7 D. 9
SOLUTION:
SOLUTION:
To find the angle, we will use apply the
Angle Between Two Lines. 5 9
Given: 𝛳 = 45°, (− 2 , − 2),
𝑚2 − 𝑚 1 y-coordinate = 6
Formula: tan 𝛳 =
1+𝑚1 𝑚2
Applying the slope-intercept form where
First we have to transform the two 5 9
𝑥 = − , 𝑦 = − , we have
equations into slope intercept form. 2 2

𝑦 = 𝑚𝑥 + 𝑏 (1)
slope − intercept form: y = mx + b
9 5
3𝑥1 + 2𝑦1 = 6 (1) − = 𝑚 (− ) + 𝑏
2 2
3
𝑦1 = − 𝑥1 + 3 𝑠𝑖𝑛𝑐𝑒 𝑚 = 𝑡𝑎𝑛 𝛳 = 45°
2
𝑥2 + 𝑦2 = 6 𝑚=1
9 5
𝑦2 = −𝑥2 + 6 − = (1) (− ) + 𝑏
2 2
3
For equation 1, the slope is − .
2 𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑏,
For equation 2, the slope is -1.
Solving for angle: 𝑏 = −2
3
−1 − (− 2) To find the x-coordinate of a point, we
tan 𝛳 =
3 have to substitute b from equation 1 and
1 + (−1)(− 2)
y-coordinate = 6. Hence,
3
−1 + 2 𝑦 = 𝑚𝑥 + 𝑏
tan 𝛳 =
3
1+2 6 = (1)𝑥 − 2

𝟏 𝒙=𝟖
𝛳 = tan−1 ( )
𝟓
𝟏
𝛳 = tan−1 ( )
𝟓
𝜭 = 𝟏𝟏. 𝟑𝟏 = 𝟏𝟏°𝟏𝟗′
Name: DONOR, Mathew S. Instructor: Engr. Marvin Radaza Practice Problems
Yr & Sec: BSEE 3 – H EE Comprehensive Course 1 - Mathematics Analytic Geometry

PROBLEM #53: The segment from (-1, 4) to PROBLEM #73: Find the angle that the line
(2)
(2,-2) is extended three times its own length. 2𝑦 − 9𝑥 − 18 = 0 makes with the x-axis.
The terminal point is
A. 74.77° C. 47.77°
A. (11, −18) C. (𝟏𝟏, −𝟐𝟎) B. 4.5° D. 𝟕𝟕. 𝟒𝟕°
B. (11, −24) D. 9
SOLUTION:
SOLUTION:
From the equation,
Let point A at (-1, 4) = (𝑥1 , 𝑥2 ), point B at
2𝑦 − 9𝑥 − 18 = 0
(2,-2) = (𝑦1, 𝑦2 ) and AC be the ratio at 1:3
𝑚1 𝑚2 1
2𝑦 = 9𝑥 + 18, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦
A B C 2
𝑚1 𝑚1 : 𝑚2 9
𝑦 = 𝑥+9
Applying the section formula at B(𝑦1, 𝑦2 ): 2
9
(𝑚1 𝑎 + 𝑚2 𝑥1 ) 𝑚=
𝑦1 = 2
𝑚1 + 𝑚2
Solving for the angle:
(𝑚1 𝑏 + 𝑚2 𝑥2 )
𝑦2 = 2𝑦 − 9𝑥
𝑚1 + 𝑚2
Substituting those value from the given, tan 𝛳 = 𝑚
1𝑎 + (3)(−1) 𝟗
𝑦1 = [ ] 𝛳 = tan−1 ( )
1+ 3 𝟐
𝑎−3 𝜭 = 𝟕𝟕. 𝟒𝟕°
2=
4
PROBLEM #93: The diameter of a circle
2(4) + 3 = 𝑎 described by 9𝑥 2 + 9𝑦 2 = 16 is:
𝒂 = 𝟏𝟏 16
A. C. 4
9
1𝑏 + (3)(4) B. 4/3 D. 8/3
𝑦2 = [ ]
1+ 3
SOLUTION:
𝑏 + 12
−2 = The diameter of the circle is twice the
4
radius. Mathematically, D = 2r. First, we
−2(4) − 12 = 𝑏 need to identify the radius based on the
𝒃 = −𝟐𝟎 given equation.

Thus, the terminal point is (11,-20) 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝐶𝑖𝑟𝑐𝑙𝑒

𝒙𝟐 + 𝒚𝟐 = 𝒓𝟐
Name: DONOR, Mathew S. Instructor: Engr. Marvin Radaza Practice Problems
Yr & Sec: BSEE 3 – H EE Comprehensive Course 1 - Mathematics Analytic Geometry

From the equation given, Standard form of the circle when the center
is not in the origin is:
9𝑥 2 + 9𝑦 2 = 16
(𝒙 − 𝒉)𝟐 + (𝒚 − 𝒌)𝟐 = 𝒓𝟐
1
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦
9 Radius of the circle is the distance from the
16 point (-4,-5) to the line 2x + 7y -10 = 0
𝑥2 + 𝑦2 =
9 2𝑥 + 7𝑦 − 10 = 0
𝑆𝑞𝑢𝑎𝑟𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 (𝑥, 𝑦) = (−4, −5)
4 4
𝑥 + 𝑦 = , 𝑤ℎ𝑒𝑟𝑒 𝑟 = 𝐴𝑥 + 𝐵𝑦 − 𝐶 = 0
3 3
[𝐴𝑥 + 𝐵𝑦 + 𝐶 ]
Solving for the diameter, D = 2r 𝑟2 =
√𝐴2 + 𝐵2
4
𝐷 = 2( ) 2(−4) + 7(−5) + (−10)
2
3 2
𝑟 =[ ]
𝟖 √22 + 72
𝑫=
𝟑
2
−53 2
𝑟 =[ ]
SET B √53
PROBLEM #13: Find the equation of the 𝑟 2 = 53
circle with the center at (-4, -5) and tangent
Finding the equation of the circle:
to the line 2𝑥 + 7𝑦 − 10 = 0.
(𝑥 − ℎ ) 2 + ( 𝑦 − 𝑘 ) 2 = 𝑟 2
A. 𝑥 2 + 𝑦 2 + 8𝑥 − 10𝑦 − 12 = 0
2 2
B. 𝑥 2 + 𝑦 2 + 8𝑥 − 10𝑦 + 12 = 0 (𝑥 − (−4)) + (𝑦 − (−5)) = 53
C. 𝒙𝟐 + 𝒚𝟐 + 𝟖𝒙 + 𝟏𝟎𝒚 − 𝟏𝟐 = 𝟎
D. 𝑥 2 + 𝑦 2 + 8𝑥 + 10𝑦 + 12 = 0 (𝑥 2 + 8𝑥 + 16) + (𝑦 2 + 10𝑦 + 25) = 53

SOLUTION: 𝑥 2 + 𝑦 2 + 8𝑥 + 10𝑦 + 41 − 53 = 0
𝒙𝟐 + 𝒚𝟐 + 𝟖𝒙 + 𝟏𝟎𝒚 − 𝟏𝟐 = 𝟎
PROBLEM #33: Find the major axis of the
ellipse 𝑥 2 + 4𝑦 2 − 2𝑥 − 8𝑦 + 1 = 0.
A. 2 C. 𝟒
B. 10 D. 6
SOLUTION:
To find the major axis of the ellipse, we
have to transform the equation into its
standard equation form of ellipse.
Name: DONOR, Mathew S. Instructor: Engr. Marvin Radaza Practice Problems
Yr & Sec: BSEE 3 – H EE Comprehensive Course 1 - Mathematics Analytic Geometry

(𝑥 − ℎ )2 (𝑦 − 𝑘 )2 𝑦2 x
+ =1 =
𝑎2 𝑏2 𝑥 2 + 𝑦 2 (𝑥 2 + 𝑦 2 )
From the problem, we have x(x 2 + 𝑦 2 )
2
𝑦 =
𝑥 2 + 4𝑦 2 − 2𝑥 − 8𝑦 + 1 = 0 (𝑥 2 + 𝑦 2 )

(𝑥 2 − 2𝑥) + 4(𝑦 2 − 2𝑦) = −1 𝑦2 = 𝑥

By completing the square, We conclude that this problem is parabola.

(𝑥 − 1)2 + 4(𝑦 − 1)2 = −1 + 1 + 4 Comparing with the standard equation:

(𝑥 − 1)2 + 4(𝑦 − 1)2 = 4 𝑦 2 = 4𝑎𝑥

Divide both sides by 4, we have Solving for a:

(𝑥 − 1)2 (𝑦 − 1)2 𝑦2 = 𝑥
+ =1
4 1 𝑦 2 = 4𝑎𝑥
𝑎2 = 4 𝑥 = 4𝑎𝑥
𝑎=2 1 = 4𝑎
Major axis, 2a = 2(2) 1
𝑎=
𝑴𝒂𝒋𝒐𝒓 𝒂𝒙𝒊𝒔 = 𝟒 4
Thus, the x & y coordinates of the focus is
PROBLEM #53: What are the x and y
𝟏
coordinates of the focus of the conic section ( , 𝟎)
𝟒
described by the following equation (Angle ϴ
corresponds to a right triangle with adjacent 𝑥2 𝑦2
PROBLEM #73: Given an ellipse + = 1.
36 32
side x,r.) 𝑟𝑠𝑖𝑛2 ϴ = cos ϴ
Determine the distance between the foci.
𝟏
A. ( , 𝟎) C. (0,0) A. 𝟐 C. 4
𝟒
𝑛 1 B. 3 D. 8
B. (0, 2 ) D. (− 2 , 0)
Based on the right triangle SOLUTION:
SOLUTION:
we can obtain equations.
To determine the distance between the foci,
𝑦
y r 𝑠𝑖𝑛ϴ = let us first determine the value of a and b.
𝑟 𝑟 2 = √ 𝑥2 + 𝑦 2
ϴ 𝑥 The standard form equation of the ellipse is
𝑐𝑜𝑠ϴ = given by
x 𝑟 𝑟 = 𝑥2 + 𝑦 2
𝑥2 𝑦2
From the given equation, + =1
𝑎2 𝑏 2
𝑟𝑠𝑖𝑛2 ϴ = cos ϴ
𝑦 2 x
(𝑥 2 + 𝑦 2 ) ( 2 2
) = 2
𝑥 +𝑦 (𝑥 + 𝑦 2 )
Name: DONOR, Mathew S. Instructor: Engr. Marvin Radaza Practice Problems
Yr & Sec: BSEE 3 – H EE Comprehensive Course 1 - Mathematics Analytic Geometry

From the equation above,

𝑥2 𝑦2 PART 2
+ =1
36 32
PROBLEM #13: What is the equation of the
Based on the given above, we now have the line that passes thru (4,0) and is parallel to
value of a and be which is 6, and 2√8, line 𝑥 − 𝑦 − 2 = 0
respectively. The distance between the foci
A. 1 C. 3
is 2c. WE can solved c from the relationship:
B. 2 D. 4
𝑎2 = 𝑏 2 + 𝑐 2
SOLUTION:
Substituting the values,

36 = 32 + 𝑐 2

𝑐 = √36 − 32
𝑐=4
(4,0)
Solving the foci, 2c
2(2) = 𝟒 𝒖𝒏𝒊𝒕𝒔
PROBLEM #93: The polar form of the
equation 3𝑥 + 4𝑦 − 2 = 0 is:
A. 3𝑟 𝑠𝑖𝑛ϴ + 4𝑟 𝑐𝑜𝑠ϴ = 2
B. 3𝑟 𝑐𝑜𝑠ϴ + 4𝑟 𝑠𝑖𝑛ϴ = −2 From the slope intercept form:
C. 𝟑𝒓 𝒄𝒐𝒔𝜭 + 𝟒𝒓 𝒔𝒊𝒏𝜭 = 𝟐
D. 3𝑟 𝑠𝑖𝑛ϴ + 4𝑟 𝑡𝑎𝑛ϴ = −2 . 𝑦 = 𝑚𝑥 + 𝑏; 𝑤ℎ𝑒𝑟𝑒 𝑏 = −2

SOLUTION: 𝑦 = 𝑥−2 (1)

The polar form of the x and y coordinate is To find the equation of the line that passes
given by 𝑥 = 𝑐𝑜𝑠ϴ & 𝑦 = 𝑠𝑖𝑛ϴ. through (4,0) and is parallel to equation,
Substituting the polar values of x & y: we will obtain the second equation

3𝑥 + 4𝑦 − 2 = 0 𝑦 = 𝑚𝑥 + 𝑐; @ 𝑥 = 4, 𝑦 = 0

3(𝑟𝑐𝑜𝑠ϴ) + 4(𝑟𝑠𝑖𝑛ϴ) − 2 = 0 0 = (1)(4) + 𝑐

𝟑𝒓 𝒄𝒐𝒔𝜭 + 𝟒𝒓 𝒔𝒊𝒏𝜭 − 𝟐 = 𝟎 𝒄 = −𝟒
Thus, the equation that passes through
(4,0) is 𝒚 = 𝒙 − 𝟒
Name: DONOR, Mathew S. Instructor: Engr. Marvin Radaza Practice Problems
Yr & Sec: BSEE 3 – H EE Comprehensive Course 1 - Mathematics Analytic Geometry

PROBLEM #33: The vertices of a triangle are PROBLEM #53: The latus rectum of an ellipse
at A(3, 4, -5), B(3, 4, 7), and C (0, 0, 0). is 6.4𝑚 long. If the semi-minor axis is 4𝑚.
Determine the length of the median for A to What is the area of the ellipse in 𝑚2 ?
side BC.
A. 60.8 𝑚2 C. 𝟔𝟐. 𝟖 𝒎𝟐
A. B. 61.8 𝑚2 D. 63.8 𝑚2
B. 5.86 C. 7.86
SOLUTION:
C. 6.86 D. 𝟖. 𝟖𝟔
2𝑏 2
SOLUTION: The latus rectum is given by, 𝑙 = ,
z 𝑎
where l = 64m and semi minor axis, b = 4m.
To find the semi major axis, we have
A B
2𝑏2
𝑙=
BC 𝑎
2(4)2
C y 𝑎=
6.4
𝑎=5
The area of the ellipse is given by:
𝐴𝑟𝑒𝑎𝑒𝑙𝑙𝑖𝑝𝑠𝑒 = 𝜋𝑎𝑏
x
𝐴𝑟𝑒𝑎𝑒𝑙𝑙𝑖𝑝𝑠𝑒 = 𝜋(5)(4)
To solve for the length of the median, we
will first solve for the midpoint at BC 𝑨𝒓𝒆𝒂𝒆𝒍𝒍𝒊𝒑𝒔𝒆 = 𝟔𝟐. 𝟖 𝒎𝟐

Given coordinates at B(3,4,7) and C(0,0,0): PROBLEM #73: The slope of the
line 3𝑥 + 2𝑦 + 5 = 0 is
Midpoint Formula:
2 3
𝑀𝐵𝐶 =
𝑥1 +𝑥2 𝑦1 + 𝑦2 𝑧1 + 𝑧2
, , A. − 3 C. 2
2 2 2
𝟑 2
𝑀𝐵𝐶
3 4 7
= , ,
B. − 𝟐 D. 3
2 2 2
3 7 SOLUTION:
𝑀𝐵𝐶 = , 2,
2 2 The slope intercept form is given by,
Solving the distance AD: 𝑦 = 𝑚𝑥 + 𝑏
𝐴𝐷 = √( 𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦2 )2 + (𝑧2 − 𝑧2 )2 𝑤ℎ𝑒𝑟𝑒 𝑚 = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒
3 7 Solving form from the equation,
𝐴𝐷 = √( − 3)2 + (2 − 4)2 + ( + 5)2
2 2
2𝑦 = −3𝑥 − 5
𝑨𝑫 = 𝟖. 𝟖𝟔
Name: DONOR, Mathew S. Instructor: Engr. Marvin Radaza Practice Problems
Yr & Sec: BSEE 3 – H EE Comprehensive Course 1 - Mathematics Analytic Geometry
1 𝑨𝒓𝒆𝒂𝒆𝒍𝒍𝒊𝒑𝒔𝒆 = 𝟏𝟓𝝅 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔
Multiply both sides by 2,

3 PROBLEM #113: Find the distance of the

𝑦 =− 𝑥−5 vertex of the curve (𝑥 − 2)2 = 4𝑦 to the line
2
𝟑
2𝑥 − 3𝑦 − 8 = 0
Therefore, 𝒎 = −
𝟐
A. 0.10 C.2.10
PROBLEM #93: The area of the ellipse B. 𝟏. 𝟏𝟎 D. 3.10
9𝑥 2 + 25𝑦 2 − 36𝑥 − 189 = 0 equal to
SOLUTION:
A. 𝟏𝟓𝝅 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 C. 25𝜋 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
The given curve (𝑥 − 2)2 = 4𝑦 is a parabola
B. 20𝜋 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠 D. 30𝜋 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
at vertex (2,0).
SOLUTION:
The equation of the line is given by
The area of the ellipse is given by,
2𝑥 − 3𝑦 − 8 = 0
𝐴𝑟𝑒𝑎𝑒𝑙𝑙𝑖𝑝𝑠𝑒 = 𝜋𝑎𝑏
The distance formula is given by:
Standard Equation Form of the Ellipse at |𝐴𝑥 + 𝐵𝑦 + 𝐶|
no origin (h,k) 𝑑=
√𝐴2 + 𝐵2
(𝑥 − ℎ )2 (𝑦 − 𝑘 )2 |2(2) − 3(0) − 8|
+ =1 𝑑=
𝑎2 𝑏2 √22 + 32
4
From the equation, 𝑑=
√13
9𝑥 2 + 25𝑦 2 − 36𝑥 − 189 = 0 𝒅 ≈ 𝟏. 𝟏𝟎
2 2
9( 𝑥 − 4𝑥) + 25𝑦 = 189 PROBLEM #133: The distance between the
By completing the square on x term: vertices of an ellipse is 10. The distance
between the foci is 6. What is the distance
9(𝑥 2 − 4𝑥 + 4) + 25𝑦 2 = 189 + 36
between the directrices?
9(𝑥 2 − 2)2 + 25𝑦 2 = 225
A. 15.7 C.17.7
1
Divide both sides by B. 𝟏𝟔. 𝟕 D. 18.7
225

9(𝑥 2 − 2)2 25𝑦 2 SOLUTION:

+ =1
25 9
The formula for directrices is given by
𝑎2 = 25 ; 𝑏2 =9
2𝑎2
𝐷𝑖𝑟𝑒𝑐𝑡𝑟𝑖𝑐𝑒𝑠 =
Hence the value for a nd b are 5 and 3, 𝑐
respectively.
Since we already have a vertex at a=5 and
Solving for the area of the ellipse: focus at c = 3, we can now obtain the b
using Pythagorean theorem.
𝐴𝑟𝑒𝑎𝑒𝑙𝑙𝑖𝑝𝑠𝑒 = 𝜋𝑎𝑏
𝑐 2 = 𝑎2 + 𝑏 2
𝐴𝑟𝑒𝑎𝑒𝑙𝑙𝑖𝑝𝑠𝑒 = 𝜋(5)(3)
Name: DONOR, Mathew S. Instructor: Engr. Marvin Radaza Practice Problems
Yr & Sec: BSEE 3 – H EE Comprehensive Course 1 - Mathematics Analytic Geometry

𝑏2 = √52 − 32 From the equation,

𝑏=4 𝑥 2 − 4𝑦 2 − 2𝑥 − 63 = 0
(𝑥 2 − 2𝑥) − 4𝑦 2 = 63
Now we obtained the value of b, we
can now write the equation of an By completing the squares on the x term:
ellipse at origin represented as (𝑥 2 − 2𝑥 + 1) − 4𝑦 2 = 63 + 1
𝑥 2 𝑦2 (𝑥 2 − 1)2 − 4𝑦 2 = 64
+ =1
𝑎2 𝑏 2 1
Dividing both sides by 64,
𝑥 2 𝑦2
+ =1 (𝑥 2 − 1)2 𝑦 2
52 42 64
−
16
= 64

𝑥 2 𝑦2 𝑎 = 8 ; 𝑏 = 4; ℎ = 1; 𝑘 = 0
+ =1
25 16 𝑏
2 𝑦 = ± (𝑥 − ℎ ) + 𝑘
Taking the value of 𝑎 to solve for 𝑎
directrices, we have 4
𝑦 = ± (𝑥 − 1) + 0
8
2𝑎2
𝐷𝑖𝑟𝑒𝑐𝑡𝑟𝑖𝑐𝑒𝑠 = 1 1
𝑐 𝑦 = ± (𝑥 − 1); 𝑚 =
2 2
2(25) Solving for the angle:
𝐷𝑖𝑟𝑒𝑐𝑡𝑟𝑖𝑐𝑒𝑠 =
3
𝑏
𝑫𝒊𝒓𝒆𝒄𝒕𝒓𝒊𝒄𝒆𝒔 = 𝟏𝟔. 𝟔𝟕 𝑚 = 𝑡𝑎𝑛ϴ =
𝑎
PROBLEM #153: What is the angle in 4
𝑡𝑎𝑛ϴ =
degrees, between an asymptote of the 8
hyperbola 𝑥 2 − 4𝑦 2 − 2𝑥 − 63 = 0 and the 1
x-axis? ϴ = tan−1 ( )
2
A. 25.6° C.27.6° 𝜭 = 𝟐𝟔. 𝟓𝟕°
B. 𝟐𝟔. 𝟔° D. 28.6°
SOLUTION:
PROBLEM #173: Using polar coordinates, a
The equation of the asymptote is given by point is at (6,70°)
𝑏 A. (−2.05, −5.64) C.(𝟐. 𝟎𝟓, 𝟓. 𝟔𝟒)
𝑦 = ± (𝑥 − ℎ ) + 𝑘
𝑎 B. (2.05, −5.64) D. (−2.05, 5.64)
Where (h,k) is the center of the hyperbola SOLUTION:
𝑏
and 𝑎 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒.
The polar coordinate given by
Name: DONOR, Mathew S. Instructor: Engr. Marvin Radaza Practice Problems
Yr & Sec: BSEE 3 – H EE Comprehensive Course 1 - Mathematics Analytic Geometry

(𝑟, ϴ)

From the given problem, we have

(6, 70°)

Take note that 𝑥 = 𝑟𝑐𝑜𝑠ϴ & 𝑦 = 𝑟𝑠𝑖𝑛ϴ

Transforming the polar coordinates into
rectangular coordinates, we obtain
𝑥 = 𝑟𝑐𝑜𝑠ϴ
𝑥 = 6cos (70)
𝑥 ≈ 2.05
𝑦 = 𝑟𝑠𝑖𝑛ϴ
𝑦 = 6sin (70)
𝑦 ≈ 5.64
Hence, the coordinates are (2.05, 5.64)
PROBLEM #193: Find the equation of the
parabola with focus at (0,8) and directrix
𝑦+8=0
A. 𝑥 2 = −32𝑦 C.𝑦 2 = −32𝑥
B. 𝒙𝟐 = 𝟑𝟐𝒚 D. 𝑦 2 = 32𝑥
SOLUTION:
Since the directrix(p) y = -8 is vertical, the
equation of the parabola opens either up or
down.
(𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘 )
@ center at the origin

𝑥 2 = 4(−8)𝑦
𝒙𝟐 = −𝟑𝟐𝒚, 𝒐𝒑𝒆𝒏𝒔 𝒖𝒑𝒘𝒂𝒓𝒅