Practice Exercise Solutions: Convert each equation to its standard form to determine the key points to trace the
graph of
the ellipse.
1. 𝟏𝟔𝒙𝟐 + 𝟑𝟔𝒚𝟐 + 𝟏𝟔𝒙 + 𝟑𝟔𝒚 − 𝟓𝟔𝟑 = 𝟎
STEP 1: Move the constant on the right-hand side of the equation and group terms with the same variable
(16𝑥 2 + 16𝑥 ) + (36𝑦 2 + 36𝑦 ) = 563
STEP 2: Reduce the coefficients of the leading terms to 1 by factoring before completing the square.
16𝑥 2 16𝑥 36𝑦 2 36𝑦
16( + ) + 36( + ) = 563 → 16(𝑥 2 + 𝑥 ) + 36(𝑦 2 + 𝑦 ) = 563
16 16 36 36
STEP 3: Complete the square for each group
𝑏 2 1 2 𝑏 2 1 2
For 𝑥 2 + 𝑥; 𝑏 = 1, ( ) = ( ) = 1/4 For 𝑦 2 + 𝑦; 𝑏 = 1, ( ) = ( ) = 1/4
2 2 2 2
STEP 4: From step 3, the constants added to each group to form perfect square trinomials must also be added to the other
side of the equation to maintain balance. Be sure to consider the constants factored out from each group. Multiply each
added constant by its corresponding factored-out coefficient (e.g., 1/4), and the result will be the values added to the
other side. Finally, simplify the constants on the right side.
1 1
16(𝑥 2 + 𝑥 + 1/4) + 36 (𝑦 2 + 𝑦 + ) = 563 + 4 + 9 → 16(𝑥 2 + 𝑥 + 1/4) + 36 (𝑦 2 + 𝑦 + ) = 576
4 4
STEP 5: Now, we can write each group as a square of binomial expression. Following the standard form of ellipse, we will
reduce the right-hand side to 1, and so we divide all terms by 576.
16(𝑥 + 1/2)2 36(𝑦 + 1/2)2 576 16(𝑥 + 1/2)2 36(𝑦 + 1/2)2
+ = → + =1
576 576 576 576 576
STEP 6: Reduce the factor of each group to 1 to come up with the standard form of the equation. For x group, divide
both the numerator and denominator by 16. And for y group, divide both the numerator and denominator by 36. Then,
simplify.
16(𝑥 + 1/2)2 /16 36(𝑦 + 1/2)2 /36 (𝑥 + 1/2)2 (𝑦 + 1/2)2
+ =1 → + =1
576/16 576/36 36 16
Take Note: Denominators can be whole numbers or rational numbers, but in both cases, they must be positive.
STEP 7: Determine the center, the distance from the center to any vertex (in 'a' units), and the distance from the center
to any co-vertex (in 'b' units).
▪ 𝑐𝑒𝑛𝑡𝑒𝑟 (−1/2, −1/2) 𝑎2 = 36 → 𝑎 = 6 𝑏 2 = 16 → 𝑏 = 4
STEP 8: Sketch the graph. Since Group X has the greater denominator, the graph has a horizontal orientation/horizontal
major axis. The value of a and b will determine the coordinates of the vertices and co-vertices.
When the Major Axis is horizontal, the coordinates of vertices and co-vertices are as follows,
𝑉1 (ℎ + 𝑎, 𝑘) → 𝑉1 (−1/2 + 6, −1/2) → 𝑉1 (11/2, −1/2) or (5.5, −0.5)
𝑉2 (ℎ − 𝑎, 𝑘) → 𝑉2 (−1/2 − 6, −1/2) → 𝑉2 (−13/2, −1/2) or (−6.5, −0.5)
𝐶𝑉1 (ℎ, 𝑘 + 𝑏) → 𝐶𝑉1 (−1/2, −1/2 + 4) → 𝐶𝑉1 (−1/2, 7/2) or (−0.5, 3.5)
𝐶𝑉2 (ℎ, 𝑘 − 𝑏) → 𝐶𝑉2 (−1/2, −1/2 − 4) → 𝐶𝑉2 (−1/2, −9/2) or (−0.5, −4.5)
2. 𝟗𝒙𝟐 + 𝟏𝟔𝒚𝟐 + 𝟏𝟖𝒙 + 𝟔𝟒𝒚 − 𝟕𝟏 = 𝟎
STEP 1: Move the constant on the right-hand side of the equation and group terms with the same variable
(9𝑥 2 + 18𝑥 ) + (16𝑦 2 + 64𝑦 ) = 71
STEP 2: Reduce the coefficients of the leading terms to 1 by factoring before completing the square.
9𝑥 2 18𝑥 16𝑦 2 64𝑦
9( + ) + 16( + ) = 71 → 9(𝑥 2 + 2𝑥 ) + 16(𝑦 2 + 4𝑦 ) = 71
9 9 16 16
STEP 3: Complete the square for each group
2 2 4 2
For 𝑥 2 + 2𝑥; 𝑏 = 2, ( ) = (1)2 = 1 For 𝑦 2 + 4𝑦; 𝑏 = 4, ( ) = (2)2 = 4
2 2
STEP 4: From step 3, the constants added to each group to form perfect square trinomials must also be added to the other
side of the equation to keep it balanced. Be mindful of the constants factored out from each group. For example, multiply
1 by 9 and 4 by 16—these products represent the actual values to be added to the other side. Then, simplify the constants
on the right side of the equation.
9(𝑥 2 + 2𝑥 + 1) + 16(𝑦 2 + 4𝑦 + 4) = 71 + 9 + 64 → 9(𝑥 2 + 2𝑥 + 1) + 16(𝑦 2 + 4𝑦 + 4) = 144
STEP 5: Now, we can write each group as a square of binomial expression. Following the standard form of ellipse, we will
reduce the right-hand side to 1, and so we divide all terms by 144.
9(𝑥 + 1)2 16(𝑦 + 2)2 144 16(𝑥 + 1)2 36(𝑦 + 2)2
+ = → + =1
144 144 144 144 144
STEP 6: Reduce the factor of each group to 1 to come up with the standard form of the equation. For x group, divide
both the numerator and denominator by 9. And for y group, divide both the numerator and denominator by 16. Then,
simplify.
9(𝑥 + 1)2 /9 16(𝑦 + 2)2 /16 (𝑥 + 1)2 (𝑦 + 2)2
+ =1 → + =1
144/9 144/16 16 9
Take Note: Denominators can be whole numbers or rational numbers, but in both cases, they must be positive.
STEP 7: Determine the center, the distance from the center to any vertex (in 'a' units), and the distance from the center
to any co-vertex (in 'b' units).
▪ 𝑐𝑒𝑛𝑡𝑒𝑟 (−1, −2) 𝑎2 = 16 → 𝑎 = 4 𝑏2 = 9 → 𝑏 = 3
STEP 8: Sketch the graph. Since Group X has the greater denominator, the graph has a horizontal orientation/horizontal
major axis. The value of a and b will determine the coordinates of the vertices and co-vertices.
When the Major Axis is horizontal, the coordinates of vertices and co-vertices are as follows,
𝑉1 (ℎ + 𝑎, 𝑘) → 𝑉1 (−1 + 4, −2) → 𝑉1 (3, −2)
𝑉2 (ℎ − 𝑎, 𝑘) → 𝑉2 (−1 − 4, −2) → 𝑉2 (−5, −2)
𝐶𝑉1 (ℎ, 𝑘 + 𝑏) → 𝐶𝑉1 (−1, −2 + 3) → 𝐶𝑉1 (−1, 1)
𝐶𝑉2 (ℎ, 𝑘 − 𝑏) → 𝐶𝑉2 (−1, −2 − 3) → 𝐶𝑉2 (−1, −5)
3. 𝒙𝟐 + 𝟑𝟔𝒚𝟐 − 𝟐𝒙 + 𝟕𝟐𝒚 + 𝟕𝟏 = 𝟎
STEP 1: Move the constant on the right-hand side of the equation and group terms with the same variable
(𝑥 2 − 2𝑥 ) + (36𝑦 2 + 72𝑦 ) = −1
STEP 2: Reduce the coefficients of the leading terms to 1 by factoring before completing the square.
36𝑦 2 72𝑦
(𝑥 2 − 2𝑥 ) + 36 ( + ) = −1 → (𝑥 2 − 2𝑥 ) + 36(𝑦 2 + 2𝑦 ) = −1
36 36
STEP 3: Complete the square for each group
−2 2 2 2
For 𝑥 2 − 2𝑥; 𝑏 = −2, ( ) = (−1)2 = 1 For 𝑦 2 + 2𝑦; 𝑏 = 2, ( ) = (1)2 = 1
2 2
STEP 4: From step 3, the constants added to each group to form perfect square trinomials must also be added to the other
side of the equation to keep it balanced. Be mindful of the constants factored out from each group. For example, multiply
1 by 36—this product represents the actual values to be added to the other side. Then, simplify the constants on the right
side of the equation.
(𝑥 2 − 2𝑥 + 1) + 36(𝑦 2 + 2𝑦 + 1) = −1 + 1 + 36 → (𝑥 2 − 2𝑥 + 1) + 36(𝑦 2 + 2𝑦 + 1) = 36
STEP 5: Now, we can write each group as a square of binomial expression. Following the standard form of ellipse, we will
reduce the right-hand side to 1, and so we divide all terms by 144.
(𝑥 − 1)2 36(𝑦 + 1)2 36 (𝑥 − 1)2 36(𝑦 + 1)2
+ = → + =1
36 36 36 36 36
STEP 6: Reduce the factor of each group to 1 to come up with the standard form of the equation. For x group, it is
already reduced. And for y group, divide both the numerator and denominator by 36. Then, simplify.
(𝑥 − 1)2 36(𝑦 + 1)2 /36 (𝑥 + 1)2 (𝑦 + 2)2
+ =1 → + =1
36 36/36 36 1
Take Note: Denominators can be whole numbers or rational numbers, but in both cases, they must be positive.
STEP 7: Determine the center, the distance from the center to any vertex (in 'a' units), and the distance from the center
to any co-vertex (in 'b' units).
▪ 𝑐𝑒𝑛𝑡𝑒𝑟 (1, −1) 𝑎2 = 36 → 𝑎 = 6 𝑏2 = 1 → 𝑏 = 1
STEP 8: Sketch the graph. Since Group X has the greater denominator, the graph has a horizontal orientation/horizontal
major axis. The value of a and b will determine the coordinates of the vertices and co-vertices.
When the Major Axis is horizontal, the coordinates of vertices and co-vertices are as follows,
𝑉1 (ℎ + 𝑎, 𝑘) → 𝑉1 (1 + 6, −1) → 𝑉1 (7, −1)
𝑉2 (ℎ − 𝑎, 𝑘) → 𝑉2 (1 − 6, −1) → 𝑉2 (−5, −1)
𝐶𝑉1 (ℎ, 𝑘 + 𝑏) → 𝐶𝑉1 (1, −1 + 1) → 𝐶𝑉1 (−1, 0)
𝐶𝑉2 (ℎ, 𝑘 − 𝑏) → 𝐶𝑉2 (1, −1 − 1) → 𝐶𝑉2 (1, −2)
4. 𝟐𝟓𝒙𝟐 + 𝟏𝟔𝒚𝟐 − 𝟑𝟐𝒚 − 𝟑𝟖𝟒 = 𝟎
STEP 1: Move the constant on the right-hand side of the equation and group terms with the same variable.
25𝑥 2 + (16𝑦 2 − 32𝑦 ) = 384
STEP 2: Reduce the coefficients of the leading terms to 1 by factoring before completing the square.
16𝑦 2 32𝑦
25𝑥 2 + 16 ( − ) = 384 → 25𝑥 2 + 16(𝑦 2 − 2𝑦 ) = 384
16 16
STEP 3: Complete the square for each group
−2 2
For 𝑦 2 − 2𝑦; 𝑏 = −2, ( ) = (−1)2 = 1
2
STEP 4: From step 3, the constants added to each group to form perfect square trinomials must also be added to the other
side of the equation to keep it balanced. Be mindful of the constants factored out from each group. For example, multiply
1 by 16—this product represents the actual values to be added to the other side. Then, simplify the constants on the right
side of the equation.
25𝑥 2 + 16(𝑦 2 − 2𝑦 + 1) = 384 + 16 → 25𝑥 2 + 16(𝑦 2 − 2𝑦 + 1) = 400
STEP 5: Now, we can write each group as a square of binomial expression. Following the standard form of ellipse, we will
reduce the right-hand side to 1, and so we divide all terms by 400.
25𝑥 2 16(𝑦 − 1)2 400 25𝑥 2 16(𝑦 + 1)2
+ = → + =1
400 400 400 400 400
STEP 6: Reduce the factor of each group to 1 to come up with the standard form of the equation. For x group, divide
both numerator and denominator by 25. And for y group, divide both the numerator and denominator by 16. Then,
simplify.
25𝑥 2 /25 16(𝑦 − 1)2 /16 𝑥 2 (𝑦 − 1)2
+ =1 → + =1
400/25 400/16 16 25
Take Note: Denominators can be whole numbers or rational numbers, but in both cases, they must be positive.
STEP 7: Determine the center, the distance from the center to any vertex (in 'a' units), and the distance from the center
to any co-vertex (in 'b' units).
▪ 𝑐𝑒𝑛𝑡𝑒𝑟 (0,1) 𝑎2 = 25 → 𝑎 = 5 𝑏 2 = 16 → 𝑏 = 4
STEP 8: Sketch the graph. Since Group Y has the greater denominator, the graph has a vertical orientation/ vertical major
axis. The value of a and b will determine the coordinates of the vertices and co-vertices.
When the Major Axis is vertical, the coordinates of vertices and co-vertices are as follows,
𝑉1 (ℎ, 𝑘 + 𝑎) → 𝑉1 (0, 1 + 5) → 𝑉1 (0, 6)
𝑉2 (ℎ, 𝑘 − 𝑏) → 𝑉2 (0, 1 − 5) → 𝑉2 (0, −4)
𝐶𝑉1 (ℎ + 𝑎, 𝑘) → 𝐶𝑉1 (0 + 4, 1) → 𝐶𝑉1 (4, 1)
𝐶𝑉2 (ℎ − 𝑎, 𝑘) → 𝐶𝑉2 (0 − 4, 1) → 𝐶𝑉2 (−4, 1)
