Ellipse worksheet

1. Area: 37.7 square units

1. Find the area: π × a × b. It is the circle formula, with the radius squared swapped for the two half-widths multiplied

   $\mathrm{Area}$$=\pi \times 4\times 3$$=37.699112$

2. The horizontal axis is the longer (major) one. Find c, how far each focus sits from the center, with the right-triangle rule

   $c$$=\sqrt{4^2-3^2}$$=2.645751$

3. Divide c by the longer half-axis to get e, the eccentricity. It shows how stretched the shape is, where 0 is a circle

   $e$$=\frac{2.645751}{4}$$=0.661438$

4. The perimeter has no everyday formula. So keep taking the usual average and the geometric average of the two half-axes until they agree. That shared value is M

   $\text{AGM(}4\text{,}3\text{)}$$=3.482028$

5. Add up the gaps left over from each round. This sum gives the exact perimeter P

   $P$$=22.103492$

6. As a check, Ramanujan's shortcut formula, π(a + b)(1 + 3h ÷ (10 + √(4 − 3h))) with h = (a − b)² ÷ (a + b)², gives almost the same answer

   $P$$=\text{≈}22.103492$

7. Write the ellipse's equation in standard form, with its center at (h, k)

   $\text{x²/16 + y²/9 = 1}$

8. The two foci sit on the longer axis, a distance c on each side of the center

   $\text{(−2.645751, 0) and (2.645751, 0)}$

2. Area: 37.7 square units

1. Find the area: π × a × b. It is the circle formula, with the radius squared swapped for the two half-widths multiplied

   $\mathrm{Area}$$=\pi \times 6\times 2$$=37.699112$

2. The horizontal axis is the longer (major) one. Find c, how far each focus sits from the center, with the right-triangle rule

   $c$$=\sqrt{6^2-2^2}$$=5.656854$

3. Divide c by the longer half-axis to get e, the eccentricity. It shows how stretched the shape is, where 0 is a circle

   $e$$=\frac{5.656854}{6}$$=0.942809$

4. The perimeter has no everyday formula. So keep taking the usual average and the geometric average of the two half-axes until they agree. That shared value is M

   $\text{AGM(}6\text{,}2\text{)}$$=3.727234$

5. Add up the gaps left over from each round. This sum gives the exact perimeter P

   $P$$=26.729786$

6. As a check, Ramanujan's shortcut formula, π(a + b)(1 + 3h ÷ (10 + √(4 − 3h))) with h = (a − b)² ÷ (a + b)², gives almost the same answer

   $P$$=\text{≈}26.729786$

7. Write the ellipse's equation in standard form, with its center at (h, k)

   $\text{x²/36 + y²/4 = 1}$

8. The two foci sit on the longer axis, a distance c on each side of the center

   $\text{(−5.656854, 0) and (5.656854, 0)}$

3. Area: 25.13 square units

1. Find the area: π × a × b. It is the circle formula, with the radius squared swapped for the two half-widths multiplied

   $\mathrm{Area}$$=\pi \times 4\times 2$$=25.132741$

2. The horizontal axis is the longer (major) one. Find c, how far each focus sits from the center, with the right-triangle rule

   $c$$=\sqrt{4^2-2^2}$$=3.464102$

3. Divide c by the longer half-axis to get e, the eccentricity. It shows how stretched the shape is, where 0 is a circle

   $e$$=\frac{3.464102}{4}$$=0.866025$

4. The perimeter has no everyday formula. So keep taking the usual average and the geometric average of the two half-axes until they agree. That shared value is M

   $\text{AGM(}4\text{,}2\text{)}$$=2.913582$

5. Add up the gaps left over from each round. This sum gives the exact perimeter P

   $P$$=19.376896$

6. As a check, Ramanujan's shortcut formula, π(a + b)(1 + 3h ÷ (10 + √(4 − 3h))) with h = (a − b)² ÷ (a + b)², gives almost the same answer

   $P$$=\text{≈}19.376896$

7. Write the ellipse's equation in standard form, with its center at (h, k)

   $\text{x²/16 + y²/4 = 1}$

8. The two foci sit on the longer axis, a distance c on each side of the center

   $\text{(−3.464102, 0) and (3.464102, 0)}$

4. Area: 62.83 square units

1. Find the area: π × a × b. It is the circle formula, with the radius squared swapped for the two half-widths multiplied

   $\mathrm{Area}$$=\pi \times 5\times 4$$=62.831853$

2. The horizontal axis is the longer (major) one. Find c, how far each focus sits from the center, with the right-triangle rule

   $c$$=\sqrt{5^2-4^2}$$=3$

3. Divide c by the longer half-axis to get e, the eccentricity. It shows how stretched the shape is, where 0 is a circle

   $e$$=\frac{3}{5}$$=0.6$

4. The perimeter has no everyday formula. So keep taking the usual average and the geometric average of the two half-axes until they agree. That shared value is M

   $\text{AGM(}5\text{,}4\text{)}$$=4.486057$

5. Add up the gaps left over from each round. This sum gives the exact perimeter P

   $P$$=28.361668$

6. As a check, Ramanujan's shortcut formula, π(a + b)(1 + 3h ÷ (10 + √(4 − 3h))) with h = (a − b)² ÷ (a + b)², gives almost the same answer

   $P$$=\text{≈}28.361668$

7. Write the ellipse's equation in standard form, with its center at (h, k)

   $\text{x²/25 + y²/16 = 1}$

8. The two foci sit on the longer axis, a distance c on each side of the center

   $\text{(−3, 0) and (3, 0)}$

5. Area: 56.55 square units

1. Find the area: π × a × b. It is the circle formula, with the radius squared swapped for the two half-widths multiplied

   $\mathrm{Area}$$=\pi \times 6\times 3$$=56.548668$

2. The horizontal axis is the longer (major) one. Find c, how far each focus sits from the center, with the right-triangle rule

   $c$$=\sqrt{6^2-3^2}$$=5.196152$

3. Divide c by the longer half-axis to get e, the eccentricity. It shows how stretched the shape is, where 0 is a circle

   $e$$=\frac{5.196152}{6}$$=0.866025$

4. The perimeter has no everyday formula. So keep taking the usual average and the geometric average of the two half-axes until they agree. That shared value is M

   $\text{AGM(}6\text{,}3\text{)}$$=4.370373$

5. Add up the gaps left over from each round. This sum gives the exact perimeter P

   $P$$=29.065345$

6. As a check, Ramanujan's shortcut formula, π(a + b)(1 + 3h ÷ (10 + √(4 − 3h))) with h = (a − b)² ÷ (a + b)², gives almost the same answer

   $P$$=\text{≈}29.065345$

7. Write the ellipse's equation in standard form, with its center at (h, k)

   $\text{x²/36 + y²/9 = 1}$

8. The two foci sit on the longer axis, a distance c on each side of the center

   $\text{(−5.196152, 0) and (5.196152, 0)}$

6. Area: 47.12 square units

1. Find the area: π × a × b. It is the circle formula, with the radius squared swapped for the two half-widths multiplied

   $\mathrm{Area}$$=\pi \times 5\times 3$$=47.12389$

2. The horizontal axis is the longer (major) one. Find c, how far each focus sits from the center, with the right-triangle rule

   $c$$=\sqrt{5^2-3^2}$$=4$

3. Divide c by the longer half-axis to get e, the eccentricity. It shows how stretched the shape is, where 0 is a circle

   $e$$=\frac{4}{5}$$=0.8$

4. The perimeter has no everyday formula. So keep taking the usual average and the geometric average of the two half-axes until they agree. That shared value is M

   $\text{AGM(}5\text{,}3\text{)}$$=3.936236$

5. Add up the gaps left over from each round. This sum gives the exact perimeter P

   $P$$=25.526999$

6. As a check, Ramanujan's shortcut formula, π(a + b)(1 + 3h ÷ (10 + √(4 − 3h))) with h = (a − b)² ÷ (a + b)², gives almost the same answer

   $P$$=\text{≈}25.526999$

7. Write the ellipse's equation in standard form, with its center at (h, k)

   $\text{x²/25 + y²/9 = 1}$

8. The two foci sit on the longer axis, a distance c on each side of the center

   $\text{(−4, 0) and (4, 0)}$

7. Area: 87.96 square units

1. Find the area: π × a × b. It is the circle formula, with the radius squared swapped for the two half-widths multiplied

   $\mathrm{Area}$$=\pi \times 7\times 4$$=87.964594$

2. The horizontal axis is the longer (major) one. Find c, how far each focus sits from the center, with the right-triangle rule

   $c$$=\sqrt{7^2-4^2}$$=5.744563$

3. Divide c by the longer half-axis to get e, the eccentricity. It shows how stretched the shape is, where 0 is a circle

   $e$$=\frac{5.744563}{7}$$=0.820652$

4. The perimeter has no everyday formula. So keep taking the usual average and the geometric average of the two half-axes until they agree. That shared value is M

   $\text{AGM(}7\text{,}4\text{)}$$=5.395248$

5. Add up the gaps left over from each round. This sum gives the exact perimeter P

   $P$$=35.203162$

6. As a check, Ramanujan's shortcut formula, π(a + b)(1 + 3h ÷ (10 + √(4 − 3h))) with h = (a − b)² ÷ (a + b)², gives almost the same answer

   $P$$=\text{≈}35.203162$

7. Write the ellipse's equation in standard form, with its center at (h, k)

   $\text{x²/49 + y²/16 = 1}$

8. The two foci sit on the longer axis, a distance c on each side of the center

   $\text{(−5.744563, 0) and (5.744563, 0)}$

8. Area: 28.27 square units

1. Find the area: π × a × b. It is the circle formula, with the radius squared swapped for the two half-widths multiplied

   $\mathrm{Area}$$=\pi \times 3\times 3$$=28.274334$

2. The horizontal axis is the longer (major) one. Find c, how far each focus sits from the center, with the right-triangle rule

   $c$$=\sqrt{3^2-3^2}$$=0$

3. Divide c by the longer half-axis to get e, the eccentricity. It shows how stretched the shape is, where 0 is a circle

   $e$$=\frac{0}{3}$$=0$

4. The perimeter has no everyday formula. So keep taking the usual average and the geometric average of the two half-axes until they agree. That shared value is M

   $\text{AGM(}3\text{,}3\text{)}$$=3$

5. Add up the gaps left over from each round. This sum gives the exact perimeter P

   $P$$=18.849556$

6. As a check, Ramanujan's shortcut formula, π(a + b)(1 + 3h ÷ (10 + √(4 − 3h))) with h = (a − b)² ÷ (a + b)², gives almost the same answer

   $P$$=\text{≈}18.849556$

7. Write the ellipse's equation in standard form, with its center at (h, k)

   $\text{x²/9 + y²/9 = 1}$

9. Area: 84.82 square units

1. Find the area: π × a × b. It is the circle formula, with the radius squared swapped for the two half-widths multiplied

   $\mathrm{Area}$$=\pi \times 9\times 3$$=84.823002$

2. The horizontal axis is the longer (major) one. Find c, how far each focus sits from the center, with the right-triangle rule

   $c$$=\sqrt{9^2-3^2}$$=8.485281$

3. Divide c by the longer half-axis to get e, the eccentricity. It shows how stretched the shape is, where 0 is a circle

   $e$$=\frac{8.485281}{9}$$=0.942809$

4. The perimeter has no everyday formula. So keep taking the usual average and the geometric average of the two half-axes until they agree. That shared value is M

   $\text{AGM(}9\text{,}3\text{)}$$=5.59085$

5. Add up the gaps left over from each round. This sum gives the exact perimeter P

   $P$$=40.09468$

6. As a check, Ramanujan's shortcut formula, π(a + b)(1 + 3h ÷ (10 + √(4 − 3h))) with h = (a − b)² ÷ (a + b)², gives almost the same answer

   $P$$=\text{≈}40.094678$

7. Write the ellipse's equation in standard form, with its center at (h, k)

   $\text{x²/81 + y²/9 = 1}$

8. The two foci sit on the longer axis, a distance c on each side of the center

   $\text{(−8.485281, 0) and (8.485281, 0)}$

10. Area: 18.85 square units

1. Find the area: π × a × b. It is the circle formula, with the radius squared swapped for the two half-widths multiplied

   $\mathrm{Area}$$=\pi \times 3\times 2$$=18.849556$

2. The horizontal axis is the longer (major) one. Find c, how far each focus sits from the center, with the right-triangle rule

   $c$$=\sqrt{3^2-2^2}$$=2.236068$

3. Divide c by the longer half-axis to get e, the eccentricity. It shows how stretched the shape is, where 0 is a circle

   $e$$=\frac{2.236068}{3}$$=0.745356$

4. The perimeter has no everyday formula. So keep taking the usual average and the geometric average of the two half-axes until they agree. That shared value is M

   $\text{AGM(}3\text{,}2\text{)}$$=2.47468$

5. Add up the gaps left over from each round. This sum gives the exact perimeter P

   $P$$=15.86544$

6. As a check, Ramanujan's shortcut formula, π(a + b)(1 + 3h ÷ (10 + √(4 − 3h))) with h = (a − b)² ÷ (a + b)², gives almost the same answer

   $P$$=\text{≈}15.86544$

7. Write the ellipse's equation in standard form, with its center at (h, k)

   $\text{x²/9 + y²/4 = 1}$

8. The two foci sit on the longer axis, a distance c on each side of the center

   $\text{(−2.236068, 0) and (2.236068, 0)}$