Derive Equations of Ellipses and Parabolas - Geometry

An ellipse has foci at $F_1(-1,2)$ and $F_2(7,2)$ . For any point $P(x,y)$ on the ellipse, $PF_1+PF_2=12$ . Which equation represents the ellipse? (No vertices are given.)
Answer: $\dfrac{(x-3)^2}{36}+\dfrac{(y-2)^2}{20}=1$ We need to derive the equation of an ellipse from its foci and distance sum. An ellipse is defined as all points P(x,y) where PF₁ + PF₂ equals a constant. The foci F₁(-1,2) and F₂(7,2) have the same y-coordinate, so the major axis is horizontal and the center is at ((−1+7)/2, 2) = (3,2). With PF₁ + PF₂ = 12, we have 2a = 12, so a = 6 and a² = 36. The distance between foci is 2c = 8, so c = 4, and b² = a² - c² = 36 - 16 = 20, giving (x-3)²/36 + (y-2)²/20 = 1. A common error is confusing which denominator is larger; remember that a > b for ellipses. Start by finding the center as the midpoint of the foci.
An ellipse has foci at $F_1(-1, 2)$ and $F_2(5, 2)$ . If point $Q(2, 6)$ lies on the ellipse, what is the equation of the ellipse in standard form centered at its center?
Answer: $\frac{(x-2)^2}{25} + \frac{(y-2)^2}{16} = 1$ 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: $(\frac{-1+5}{2}, \frac{2+2}{2}) = (2, 2)$ . Since both foci have the same y-coordinate, this is a horizontal ellipse. The distance between foci is $|5-(-1)| = 6$ , so $2c = 6$ and $c = 3$ . Now calculate the sum of distances from point $Q(2, 6)$ to both foci: - Distance to $F_1(-1, 2)$ : $\sqrt{(2-(-1))^2 + (6-2)^2} = \sqrt{9+16} = 5$ - Distance to $F_2(5, 2)$ : $\sqrt{(2-5)^2 + (6-2)^2} = \sqrt{9+16} = 5$ The sum is $2a = 10$ , so $a = 5$ . Using the relationship $c^2 = a^2 - b^2$ : $9 = 25 - b^2$ , which gives $b^2 = 16$ and $b = 4$ . The standard form is $\frac{(x-2)^2}{25} + \frac{(y-2)^2}{16} = 1$ , 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 $a^2$ under the x-term.
A hyperbola has foci $F_1(0,-10)$ and $F_2(0,10)$ , and the distance condition is $|PF_2-PF_1|=12$ for any point $P(x,y)$ on the hyperbola. Which equation represents the hyperbola? (No asymptotes are given.)
Answer: $\dfrac{y^2}{36}-\dfrac{x^2}{64}=1$ This problem asks for a vertical hyperbola equation from the focus definition. A hyperbola consists of points P(x,y) where the absolute difference of distances to foci is constant. The foci F₁(0,-10) and F₂(0,10) are vertical with center at (0,0), and c = 10. The condition |PF₂ - PF₁| = 12 gives 2a = 12, so a = 6. Using c² = a² + b² for hyperbolas, we get 100 = 36 + b², so b² = 64. Since the foci are vertical, the transverse axis is vertical, making the equation y²/a² - x²/b² = 1, which gives y²/36 - x²/64 = 1. A common error is using the wrong orientation - vertical foci mean the y² term is positive and comes first. To derive hyperbola equations systematically, always identify the transverse axis direction from the foci arrangement.
A hyperbola has foci at $F_1(-4,3)$ and $F_2(4,3)$ . For any point $P(x,y)$ on the hyperbola, $|PF_2-PF_1|=4$ . Which equation represents the hyperbola? (No asymptotes are given.)
Answer: $\dfrac{x^2}{4}-\dfrac{(y-3)^2}{12}=1$ This problem requires deriving a hyperbola equation from its geometric definition. A hyperbola is all points P(x,y) where |PF₂ - PF₁| equals a constant. The foci F₁(-4,3) and F₂(4,3) have the same y-coordinate, creating a horizontal hyperbola centered at (0,3). Given |PF₂ - PF₁| = 4, we have 2a = 4, so a = 2 and a² = 4. The focal distance is 2c = 8, so c = 4, and b² = c² - a² = 16 - 4 = 12, giving x²/4 - (y-3)²/12 = 1. Students often confuse the order of terms; for horizontal hyperbolas, x² comes first with a positive coefficient. Start by identifying the transverse axis from the foci alignment.
An ellipse is defined as the set of all points $P(x,y)$ such that the sum of the distances from $P$ to the two foci is constant. The foci are $F_1(-2,0)$ and $F_2(2,0)$ , and the constant sum of distances is $10$ . Which equation represents the ellipse? (No directrix is given.)
Answer: $\dfrac{x^2}{25}+\dfrac{y^2}{21}=1$ This problem asks us to derive the equation of an ellipse from its geometric definition. An ellipse is the set of all points P(x,y) where the sum of distances to two fixed points (foci) is constant. The foci are F₁(-2,0) and F₂(2,0), and the center is at their midpoint (0,0). The distance condition PF₁ + PF₂ = 10 means 2a = 10, so a = 5. Since c = 2 (distance from center to focus) and a² = b² + c², we get 25 = b² + 4, so b² = 21. The standard form for a horizontal ellipse centered at origin is x²/a² + y²/b² = 1, giving us x²/25 + y²/21 = 1. A common error is confusing which denominator goes with which variable - remember that a corresponds to the direction of the foci. To derive equations systematically, always start with the distance condition and work through the algebra to eliminate radicals.
An ellipse has foci at $(-2, 1)$ and $(4, 1)$ and passes through the point $(1, 5)$ . What is the value of $a^2$ in the standard form equation of this ellipse?
Answer: 25 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.
An ellipse has foci at $F_1(0,-2)$ and $F_2(0,2)$ . The sum of distances from any point $P(x,y)$ on the ellipse to the foci is $10$ . Which equation represents the ellipse? (No vertices are given.)
Answer: $\dfrac{x^2}{21}+\dfrac{y^2}{25}=1$ 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.
A parabola has focus at $(2, 5)$ and directrix $y = 1$ . Which equation represents this parabola?
Answer: $(x - 2)^2 = 8(y - 3)$ The vertex is midway between focus (2,5) and directrix y = 1, so vertex is at (2,3). The distance from vertex to focus is p = 2. Since the focus is above the directrix, the parabola opens upward with form (x-h)² = 4p(y-k). Substituting: (x-2)² = 4(2)(y-3) = 8(y-3). Choice B uses 4p = 16 instead of 8. Choice C uses the wrong orientation (horizontal instead of vertical). Choice D uses p = 1 instead of p = 2.
An ellipse has foci at $F_1(1,3)$ and $F_2(9,3)$ . For any point $P(x,y)$ on the ellipse, $PF_1+PF_2=14$ . Which equation represents the ellipse? (No vertices are given directly.)
Answer: $\dfrac{(x-5)^2}{49}+\dfrac{(y-3)^2}{33}=1$ This problem involves deriving an ellipse equation with a shifted center. An ellipse is defined by the constant sum of distances from any point to two foci. The foci F₁(1,3) and F₂(9,3) are horizontal with center at their midpoint (5,3). The distance PF₁ + PF₂ = 14 gives us 2a = 14, so a = 7. The distance between foci is 8, so 2c = 8 and c = 4. Using a² = b² + c², we get 49 = b² + 16, so b² = 33. For a horizontal ellipse centered at (h,k), the equation is (x-h)²/a² + (y-k)²/b² = 1, giving us (x-5)²/49 + (y-3)²/33 = 1. A common error is placing the center at one focus instead of between them. Always find the center as the midpoint of the foci before writing the shifted equation.
A hyperbola has foci at $F_1(-6,-1)$ and $F_2(2,-1)$ . For any point $P(x,y)$ on the hyperbola, $|PF_2-PF_1|=6$ . Which equation represents the hyperbola? (No asymptotes are given.)
Answer: $\dfrac{(x+2)^2}{9}-\dfrac{(y+1)^2}{7}=1$ This problem asks for the equation of a hyperbola given its foci and distance difference. A hyperbola consists of all points P(x,y) where |PF₂ - PF₁| is constant. The foci F₁(-6,-1) and F₂(2,-1) share the same y-coordinate, making this a horizontal hyperbola centered at ((-6+2)/2, -1) = (-2,-1). Given |PF₂ - PF₁| = 6, we have 2a = 6, so a = 3 and a² = 9. The focal distance is 2c = 8, so c = 4, and for hyperbolas b² = c² - a² = 16 - 9 = 7, yielding (x+2)²/9 - (y+1)²/7 = 1. Students often mistakenly add the squared terms instead of subtracting. Remember that hyperbolas have a difference of squares in their standard form.
An ellipse has foci at $F_1(-1,4)$ and $F_2(7,4)$ . For any point $P(x,y)$ on the ellipse, $PF_1+PF_2=18$ . Which equation follows from the focus definition? (No directrix is given.)
Answer: $\dfrac{(x-3)^2}{81}+\dfrac{(y-4)^2}{65}=1$ 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.
A parabola has vertex at $(-1, 3)$ and passes through the point $(3, 7)$ . If the parabola opens upward, what is the equation of its directrix?
Answer: $y = 2$ Since the parabola opens upward with vertex (-1,3), the equation has form (x+1)² = 4p(y-3). Using point (3,7): (3+1)² = 4p(7-3), so 16 = 16p, giving p = 1. The directrix is p units below the vertex, so it's at y = 3 - 1 = 2. Choice B places the directrix too far below the vertex. Choice C places it above the vertex. Choice D places it even further above the vertex.
A hyperbola has foci at $F_1(0,-6)$ and $F_2(0,6)$ . For any point $P(x,y)$ on the hyperbola, $|PF_2-PF_1|=8$ . Which equation represents the hyperbola? (No asymptotes are given.)
Answer: $\dfrac{y^2}{16}-\dfrac{x^2}{20}=1$ This problem asks for a vertical hyperbola equation from the focus definition. A hyperbola consists of points P(x,y) where the absolute difference of distances to two foci is constant. The foci F₁(0,-6) and F₂(0,6) are vertical with center at (0,0), and c = 6. The condition |PF₂ - PF₁| = 8 means 2a = 8, so a = 4. For hyperbolas, c² = a² + b², giving 36 = 16 + b², so b² = 20. Since the foci are vertical, the transverse axis is vertical, and the equation is y²/a² - x²/b² = 1, yielding y²/16 - x²/20 = 1. A common mistake is using the horizontal form x²/a² - y²/b² = 1 when foci are vertical. To avoid errors, identify the direction of the transverse axis from the foci positions before writing the equation.
An ellipse has foci at $F_1(-3, 0)$ and $F_2(3, 0)$ . If the point $(0, 4)$ lies on the ellipse, what is the equation of the ellipse in standard form?
Answer: $\frac{x^2}{25} + \frac{y^2}{16} = 1$ 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.
An ellipse has foci $F_1(-2,5)$ and $F_2(6,5)$ . For every point $P$ on the ellipse, $PF_1+PF_2=18$ . Which equation represents the ellipse?
Answer: $\dfrac{(x-2)^2}{81}+\dfrac{(y-5)^2}{65}=1$ To derive this ellipse equation, we use the definition: all points where the sum of distances to foci is constant. The foci F₁(-2,5) and F₂(6,5) have the same y-coordinate, creating a horizontal ellipse centered at ((-2+6)/2,5)=(2,5). The distance between foci is 2c=8, so c=4, and PF₁+PF₂=18 gives 2a=18, so a=9. From c²=a²-b², we get 16=81-b², so b²=65. For a horizontal ellipse centered at (h,k), we use (x-h)²/a²+(y-k)²/b²=1, giving (x-2)²/81+(y-5)²/65=1. Students might place 81 under the y-term, but since foci are horizontal, a² must be under the x-term. Verify by confirming the center is equidistant from both foci and that c<a for ellipse validity.
An ellipse is defined by the distance condition $PF_1+PF_2=12$ , where $F_1(-3,0)$ and $F_2(3,0)$ are the foci. Which equation represents the ellipse? (No directrix is given.)
Answer: $\dfrac{x^2}{36}+\dfrac{y^2}{27}=1$ This problem requires deriving an ellipse equation from the distance sum condition. An ellipse is the locus of points P(x,y) where PF₁ + PF₂ equals a constant. With foci F₁(-3,0) and F₂(3,0) at center (0,0), and PF₁ + PF₂ = 12, we have 2a = 12, so a = 6. The focal distance is c = 3 (from center to each focus). Using the ellipse relationship a² = b² + c², we get 36 = b² + 9, so b² = 27. The standard form for a horizontal ellipse is x²/a² + y²/b² = 1, giving x²/36 + y²/27 = 1. A common error is confusing a (semi-major axis) with c (focal distance). Remember that a is always larger than c for ellipses, and comes from the distance sum condition.
The equation of an ellipse is $\frac{(x-2)^2}{36} + \frac{(y+1)^2}{20} = 1$ . What is the distance between the two foci of this ellipse?
Answer: 8 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.
A hyperbola is defined as the set of all points $P(x,y)$ such that the absolute difference of distances to the foci $F_1(-2,0)$ and $F_2(6,0)$ is constant. Here, $|PF_2-PF_1|=2$ . Which equation follows from this focus definition? (No asymptotes are given.)
Answer: $\dfrac{(x-2)^2}{1}-\dfrac{y^2}{15}=1$ This problem requires deriving a hyperbola equation from its focus definition. A hyperbola consists of all points P(x,y) where |PF₂ - PF₁| is constant. The foci F₁(-2,0) and F₂(6,0) lie on the x-axis, creating a horizontal hyperbola centered at ((-2+6)/2, 0) = (2,0). Given |PF₂ - PF₁| = 2, we have 2a = 2, so a = 1 and a² = 1. The focal distance is 2c = 8, so c = 4, and b² = c² - a² = 16 - 1 = 15, giving (x-2)²/1 - y²/15 = 1. Students might incorrectly use b² = 12 by confusing formulas; always use b² = c² - a² for hyperbolas. Start from the geometric definition to avoid formula confusion.
An ellipse is defined as the set of points $P(x,y)$ such that $PF_1+PF_2$ is constant. The foci are $F_1(0,-4)$ and $F_2(0,4)$ , and $PF_1+PF_2=12$ . Which equation represents the ellipse?
Answer: $\dfrac{y^2}{36}+\dfrac{x^2}{20}=1$ To derive this ellipse equation, we apply the definition: points where the sum of distances to two foci equals a constant. The foci F₁(0,-4) and F₂(0,4) lie on the y-axis with center at origin, and c=4. Given PF₁+PF₂=12, we have 2a=12, so a=6. Using the ellipse relationship c²=a²-b², we get 16=36-b², so b²=20. Since foci are vertical, this is a vertical ellipse with form y²/a²+x²/b²=1, yielding y²/36+x²/20=1. Students often write x²/36+y²/20=1 by habit, but the larger denominator (a²=36) must go under the variable matching the foci direction (y here). To verify, check that any point satisfying the equation maintains the sum of distances equal to 12.
A hyperbola has foci $F_1(-1,-6)$ and $F_2(-1,2)$ . For every point $P$ on the hyperbola, $|PF_2-PF_1|=4$ . Which equation represents the hyperbola?
Answer: $\dfrac{(y+2)^2}{4}-\dfrac{(x+1)^2}{12}=1$ This hyperbola problem starts with identifying the geometric constraint: the absolute difference of distances to two foci is constant. The foci F₁(-1,-6) and F₂(-1,2) share x-coordinate -1, making this a vertical hyperbola centered at (-1,(-6+2)/2)=(-1,-2). The focal distance is 2c=8, so c=4, and |PF₂-PF₁|=4 gives 2a=4, so a=2. For hyperbolas, c²=a²+b², yielding 16=4+b², so b²=12. The vertical hyperbola form is (y-k)²/a²-(x-h)²/b²=1, giving (y+2)²/4-(x+1)²/12=1. A common error is using the wrong sign pattern or swapping a² and b² positions, but remember that a² goes under the positive term matching the foci orientation. Always verify that c>a for hyperbolas, confirming our focal geometry.

Start this quiz