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Geometry

Geometry Help: Derive Equations Of Ellipses And Parabolas

Review real example questions for Derive Equations Of Ellipses And Parabolas in Geometry.

Question 1

The equation of an ellipse is (x−2)236+(y+1)220=1\frac{(x-2)^2}{36} + \frac{(y+1)^2}{20} = 136(x−2)2​+20(y+1)2​=1. What is the distance between the two foci of this ellipse?

  1. 4
  2. 8
  3. 16
  4. 32
Explanation: From the equation, a² = 36 and b² = 20, so a = 6 and b = 2√5. Since a > b, the major axis is horizontal. Using c² = a² - b² = 36 - 20 = 16, we get c = 4. The distance between foci is 2c = 2(4) = 8. Choice A gives c instead of 2c. Choice C uses a² - b² = 16 as the distance. Choice D uses 2(a² - b²) incorrectly.

Question 2

An ellipse has foci at (−2,1)(-2, 1)(−2,1) and (4,1)(4, 1)(4,1) and passes through the point (1,5)(1, 5)(1,5). What is the value of a2a^2a2 in the standard form equation of this ellipse?

  1. 18
  2. 25
  3. 36
  4. 45
Explanation: The sum of distances from point (1,5) to both foci equals 2a. Distance from (1,5) to (-2,1) is √[(1-(-2))² + (5-1)²] = √(9+16) = 5. Distance from (1,5) to (4,1) is √[(1-4)² + (5-1)²] = √(9+16) = 5. Therefore 2a = 5 + 5 = 10, so a = 5 and a² = 25. Choice A gives a² = 18 (using incorrect calculation). Choice C uses a = 6 instead of a = 5. Choice D incorrectly uses the sum of squared distances instead of sum of distances.

Question 3

An ellipse has center at the origin and passes through points (5,0)(5, 0)(5,0) and (0,3)(0, 3)(0,3). If the foci lie on the x-axis, what are the coordinates of the foci?

  1. (±2,0)(\pm 2, 0)(±2,0)
  2. (±16,0)(\pm \sqrt{16}, 0)(±16​,0)
  3. (±34,0)(\pm \sqrt{34}, 0)(±34​,0)
  4. (±4,0)(\pm 4, 0)(±4,0)
Explanation: When you encounter an ellipse problem with given points and foci locations, you need to identify the semi-major axis aaa, semi-minor axis bbb, and then use the relationship c2=a2−b2c^2 = a^2 - b^2c2=a2−b2 to find the focal distance ccc. Since the ellipse is centered at the origin with foci on the x-axis, it has the standard form x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1a2x2​+b2y2​=1, where a>ba > ba>b. The point (5,0)(5, 0)(5,0) tells us the ellipse extends 5 units along the x-axis, so a=5a = 5a=5. The point (0,3)(0, 3)(0,3) tells us it extends 3 units along the y-axis, so b=3b = 3b=3. Now you can find the focal distance: c2=a2−b2=25−9=16c^2 = a^2 - b^2 = 25 - 9 = 16c2=a2−b2=25−9=16, so c=4c = 4c=4. The foci are located at (±c,0)=(±4,0)(\pm c, 0) = (\pm 4, 0)(±c,0)=(±4,0), which is answer D. Looking at the wrong answers: A gives (±2,0)(\pm 2, 0)(±2,0), which would result from incorrectly calculating c2=9−25=−16c^2 = 9 - 25 = -16c2=9−25=−16 and taking c=2c = 2c=2. B shows (±16,0)(\pm \sqrt{16}, 0)(±16​,0), which equals (±4,0)(\pm 4, 0)(±4,0) but leaves the square root unsimplified—this might tempt you if you forgot to simplify 16=4\sqrt{16} = 416​=4. C gives (±34,0)(\pm \sqrt{34}, 0)(±34​,0), which comes from incorrectly adding a2+b2=25+9=34a^2 + b^2 = 25 + 9 = 34a2+b2=25+9=34 instead of subtracting. Remember: for ellipses, always verify that a>ba > ba>b when foci lie on the x-axis, and use c2=a2−b2c^2 = a^2 - b^2c2=a2−b2, not addition.

Question 4

An ellipse has foci at F1(−3,0)F_1(-3, 0)F1​(−3,0) and F2(3,0)F_2(3, 0)F2​(3,0). If the point (0,4)(0, 4)(0,4) lies on the ellipse, what is the equation of the ellipse in standard form?

  1. x225+y216=1\frac{x^2}{25} + \frac{y^2}{16} = 125x2​+16y2​=1
  2. x216+y225=1\frac{x^2}{16} + \frac{y^2}{25} = 116x2​+25y2​=1
  3. x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 125x2​+9y2​=1
  4. x29+y225=1\frac{x^2}{9} + \frac{y^2}{25} = 19x2​+25y2​=1
Explanation: For an ellipse, the sum of distances from any point to the two foci is constant. Using point (0,4): distance from (0,4) to (-3,0) is √(9+16) = 5, and distance from (0,4) to (3,0) is √(9+16) = 5. So 2a = 10, thus a = 5. Since c = 3 (distance from center to focus), we have b² = a² - c² = 25 - 9 = 16, so b = 4. The equation is x²/25 + y²/16 = 1. Choice B swaps a and b values. Choice C uses b² = 9 instead of 16. Choice D incorrectly uses a = 3 and b = 5.

Question 5

An ellipse has foci at F1(−1,4)F_1(-1,4)F1​(−1,4) and F2(7,4)F_2(7,4)F2​(7,4). For any point P(x,y)P(x,y)P(x,y) on the ellipse, PF1+PF2=18PF_1+PF_2=18PF1​+PF2​=18. Which equation follows from the focus definition? (No directrix is given.)

  1. (x−3)281+(y−4)265=1\dfrac{(x-3)^2}{81}+\dfrac{(y-4)^2}{65}=181(x−3)2​+65(y−4)2​=1
  2. (x−3)265+(y−4)281=1\dfrac{(x-3)^2}{65}+\dfrac{(y-4)^2}{81}=165(x−3)2​+81(y−4)2​=1
  3. (x−3)281−(y−4)265=1\dfrac{(x-3)^2}{81}-\dfrac{(y-4)^2}{65}=181(x−3)2​−65(y−4)2​=1
  4. (x+1)281+(y−4)265=1\dfrac{(x+1)^2}{81}+\dfrac{(y-4)^2}{65}=181(x+1)2​+65(y−4)2​=1
Explanation: This problem involves deriving an ellipse equation with both horizontal and vertical shifts. An ellipse is defined by points P(x,y) where PF₁ + PF₂ equals a constant. The foci F₁(-1,4) and F₂(7,4) are horizontal with center at (3,4), and the distance between foci is 8, so c = 4. The condition PF₁ + PF₂ = 18 gives 2a = 18, so a = 9. Using a² = b² + c² for ellipses, we get 81 = b² + 16, so b² = 65. For a horizontal ellipse centered at (h,k), the equation is (x-h)²/a² + (y-k)²/b² = 1, yielding (x-3)²/81 + (y-4)²/65 = 1. A common mistake is incorrectly calculating the center - it must be the midpoint of the foci. Always verify your center coordinates before writing the shifted equation form.

Question 6

An ellipse has foci at F1(−1,2)F_1(-1, 2)F1​(−1,2) and F2(5,2)F_2(5, 2)F2​(5,2). If point Q(2,6)Q(2, 6)Q(2,6) lies on the ellipse, what is the equation of the ellipse in standard form centered at its center?

  1. (x−2)225+(y−2)29=1\frac{(x-2)^2}{25} + \frac{(y-2)^2}{9} = 125(x−2)2​+9(y−2)2​=1
  2. (x−2)216+(y−2)225=1\frac{(x-2)^2}{16} + \frac{(y-2)^2}{25} = 116(x−2)2​+25(y−2)2​=1
  3. (x−2)225+(y−2)216=1\frac{(x-2)^2}{25} + \frac{(y-2)^2}{16} = 125(x−2)2​+16(y−2)2​=1
  4. (x−2)2100+(y−2)291=1\frac{(x-2)^2}{100} + \frac{(y-2)^2}{91} = 1100(x−2)2​+91(y−2)2​=1
Explanation: When you encounter an ellipse problem with given foci and a point on the ellipse, you need to use the fundamental definition: an ellipse is the set of all points where the sum of distances to two foci is constant. First, find the center by averaging the foci coordinates: (−1+52,2+22)=(2,2)(\frac{-1+5}{2}, \frac{2+2}{2}) = (2, 2)(2−1+5​,22+2​)=(2,2). Since both foci have the same y-coordinate, this is a horizontal ellipse. The distance between foci is ∣5−(−1)∣=6|5-(-1)| = 6∣5−(−1)∣=6, so 2c=62c = 62c=6 and c=3c = 3c=3. Now calculate the sum of distances from point Q(2,6)Q(2, 6)Q(2,6) to both foci:
  • Distance to F1(−1,2)F_1(-1, 2)F1​(−1,2): (2−(−1))2+(6−2)2=9+16=5\sqrt{(2-(-1))^2 + (6-2)^2} = \sqrt{9+16} = 5(2−(−1))2+(6−2)2​=9+16​=5
  • Distance to F2(5,2)F_2(5, 2)F2​(5,2): (2−5)2+(6−2)2=9+16=5\sqrt{(2-5)^2 + (6-2)^2} = \sqrt{9+16} = 5(2−5)2+(6−2)2​=9+16​=5
The sum is 2a=102a = 102a=10, so a=5a = 5a=5. Using the relationship c2=a2−b2c^2 = a^2 - b^2c2=a2−b2: 9=25−b29 = 25 - b^29=25−b2, which gives b2=16b^2 = 16b2=16 and b=4b = 4b=4. The standard form is (x−2)225+(y−2)216=1\frac{(x-2)^2}{25} + \frac{(y-2)^2}{16} = 125(x−2)2​+16(y−2)2​=1, which is answer C. Answer A incorrectly swaps the denominators, placing 9 under the y-term instead of 16. Answer B mistakenly puts the larger denominator (25) under the y-term, suggesting a vertical ellipse when it should be horizontal. Answer D uses incorrect values (100 and 91) that don't follow from the given information. Remember: always verify your ellipse orientation by checking which foci coordinate varies—horizontal variation means horizontal ellipse with a2a^2a2 under the x-term.

Question 7

An ellipse has foci at F1(0,−2)F_1(0,-2)F1​(0,−2) and F2(0,2)F_2(0,2)F2​(0,2). The sum of distances from any point P(x,y)P(x,y)P(x,y) on the ellipse to the foci is 101010. Which equation represents the ellipse? (No vertices are given.)

  1. x225+y221=1\dfrac{x^2}{25}+\dfrac{y^2}{21}=125x2​+21y2​=1
  2. x221+y225=1\dfrac{x^2}{21}+\dfrac{y^2}{25}=121x2​+25y2​=1
  3. x225−y221=1\dfrac{x^2}{25}-\dfrac{y^2}{21}=125x2​−21y2​=1
  4. x221+(y−2)225=1\dfrac{x^2}{21}+\dfrac{(y-2)^2}{25}=121x2​+25(y−2)2​=1
Explanation: This problem asks for a vertical ellipse equation from the focus definition. An ellipse consists of points P(x,y) where the sum of distances to two foci is constant. The foci F₁(0,-2) and F₂(0,2) are vertical with center at (0,0), and c = 2. The condition PF₁ + PF₂ = 10 gives 2a = 10, so a = 5. Using a² = b² + c² for ellipses, we get 25 = b² + 4, so b² = 21. Since the foci are vertical, the major axis is vertical, making the equation x²/b² + y²/a² = 1, which gives x²/21 + y²/25 = 1. A common error is always putting a² under x² regardless of orientation - remember that a² goes with the direction of the major axis (foci direction). To derive equations correctly, identify the orientation from the foci first.

Question 8

A parabola opening rightward has vertex at (−2,4)(-2, 4)(−2,4) and focus at (1,4)(1, 4)(1,4). What is the equation of this parabola?

  1. (x+2)2=12(y−4)(x + 2)^2 = 12(y - 4)(x+2)2=12(y−4)
  2. (y−4)2=6(x+2)(y - 4)^2 = 6(x + 2)(y−4)2=6(x+2)
  3. (y−4)2=12(x+2)(y - 4)^2 = 12(x + 2)(y−4)2=12(x+2)
  4. (y−4)2=3(x+2)(y - 4)^2 = 3(x + 2)(y−4)2=3(x+2)
Explanation: When you encounter a parabola problem with vertex and focus information, you need to identify the parabola's orientation and use the appropriate standard form. Since this parabola opens rightward (horizontal orientation), you'll use the form (y−k)2=4p(x−h)(y - k)^2 = 4p(x - h)(y−k)2=4p(x−h) where (h,k)(h,k)(h,k) is the vertex and ppp is the distance from vertex to focus. The vertex is at (−2,4)(-2, 4)(−2,4), so h=−2h = -2h=−2 and k=4k = 4k=4. The focus is at (1,4)(1, 4)(1,4). Since both points have the same y-coordinate and the focus is to the right of the vertex, this confirms rightward opening. The distance ppp from vertex to focus is 1−(−2)=31 - (-2) = 31−(−2)=3. Substituting into the standard form: (y−4)2=4(3)(x−(−2))(y - 4)^2 = 4(3)(x - (-2))(y−4)2=4(3)(x−(−2)), which simplifies to (y−4)2=12(x+2)(y - 4)^2 = 12(x + 2)(y−4)2=12(x+2). Option A uses the wrong standard form (x+2)2=12(y−4)(x + 2)^2 = 12(y - 4)(x+2)2=12(y−4), which describes a parabola opening upward, not rightward. Option B has (y−4)2=6(x+2)(y - 4)^2 = 6(x + 2)(y−4)2=6(x+2), using the correct form but with 4p=64p = 64p=6 instead of 4p=124p = 124p=12. This represents a parabola with p=1.5p = 1.5p=1.5, placing the focus at (0.5,4)(0.5, 4)(0.5,4) instead of (1,4)(1, 4)(1,4). Option D also uses the correct orientation but has 4p=34p = 34p=3, giving p=0.75p = 0.75p=0.75 and placing the focus at (−1.25,4)(-1.25, 4)(−1.25,4). Remember: for horizontal parabolas, use (y−k)2=4p(x−h)(y - k)^2 = 4p(x - h)(y−k)2=4p(x−h), and always calculate ppp as the actual distance between vertex and focus to get 4p4p4p correct.

Question 9

A parabola has focus at (3,−2)(3, -2)(3,−2) and vertex at (3,1)(3, 1)(3,1). Which point lies on the directrix of this parabola?

  1. (3,−5)(3, -5)(3,−5)
  2. (3,1)(3, 1)(3,1)
  3. (3,−2)(3, -2)(3,−2)
  4. (0,4)(0, 4)(0,4)
Explanation: When you encounter a parabola problem involving focus and vertex, remember that the directrix is always positioned so that it's equidistant from the vertex as the focus, but on the opposite side. First, let's find the directrix. The vertex is at (3,1)(3, 1)(3,1) and the focus is at (3,−2)(3, -2)(3,−2). Since both points have the same x-coordinate, this is a vertical parabola. The distance from vertex to focus is ∣1−(−2)∣=3|1 - (-2)| = 3∣1−(−2)∣=3 units downward from the vertex. The directrix must be 3 units upward from the vertex, placing it at y=1+3=4y = 1 + 3 = 4y=1+3=4. So the directrix is the horizontal line y=4y = 4y=4. Now let's check which point lies on this line. Point D (0,4)(0, 4)(0,4) has a y-coordinate of 4, so it lies on the directrix y=4y = 4y=4. Why the other answers are wrong: Point A (3,−5)(3, -5)(3,−5) lies on the vertical line through the vertex and focus, but below the focus. Point B (3,1)(3, 1)(3,1) is the vertex itself, not on the directrix. Point C (3,−2)(3, -2)(3,−2) is the focus, which by definition cannot be on the directrix. Study tip: For any parabola, the vertex is always the midpoint between the focus and directrix. Once you know the focus and vertex, you can find the directrix by reflecting the focus across the vertex. Remember that the directrix is always a line perpendicular to the axis of symmetry.

Question 10

A hyperbola has foci F1(−6,3)F_1(-6,3)F1​(−6,3) and F2(2,3)F_2(2,3)F2​(2,3). For every point PPP on the hyperbola, ∣PF2−PF1∣=6|PF_2-PF_1|=6∣PF2​−PF1​∣=6. Which equation represents the hyperbola?

  1. (x+2)29−(y−3)27=1\dfrac{(x+2)^2}{9}-\dfrac{(y-3)^2}{7}=19(x+2)2​−7(y−3)2​=1
  2. (x+2)27−(y−3)29=1\dfrac{(x+2)^2}{7}-\dfrac{(y-3)^2}{9}=17(x+2)2​−9(y−3)2​=1
  3. (y−3)29−(x+2)27=1\dfrac{(y-3)^2}{9}-\dfrac{(x+2)^2}{7}=19(y−3)2​−7(x+2)2​=1
  4. (x−2)29−(y−3)27=1\dfrac{(x-2)^2}{9}-\dfrac{(y-3)^2}{7}=19(x−2)2​−7(y−3)2​=1
Explanation: This hyperbola derivation begins with the geometric definition: points where the absolute difference of distances to foci is constant. The foci F₁(-6,3) and F₂(2,3) share y-coordinate 3, indicating a horizontal hyperbola centered at ((-6+2)/2,3)=(-2,3). The focal distance is 2c=8, so c=4, and |PF₂-PF₁|=6 gives 2a=6, so a=3. Using c²=a²+b² for hyperbolas, we get 16=9+b², so b²=7. The horizontal hyperbola form is (x-h)²/a²-(y-k)²/b²=1, yielding (x+2)²/9-(y-3)²/7=1. A common mistake is using the vertical form or incorrect center, but foci alignment determines orientation. Always verify that the center-to-focus distance equals c and that c>a for valid hyperbola geometry.