This problem requires deriving a hyperbola equation from its geometric definition. A hyperbola is the set of all points where the absolute difference of distances to two foci is constant. The foci F₁(0,-5) and F₂(0,5) lie on the y-axis with center at origin and c=5, and |PF₂-PF₁|=6 means 2a=6, so a=3. For a hyperbola, c²=a²+b², giving 25=9+b², so b²=16. Since the foci are on the y-axis, this is a vertical hyperbola with equation y²/a²-x²/b²=1, yielding y²/9-x²/16=1. A common mistake is using the horizontal form x²/a²-y²/b²=1, but the foci location determines orientation. Always start by identifying whether foci align vertically or horizontally to choose the correct standard form.

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.

This hyperbola problem requires applying the geometric definition: points where the absolute difference of distances to foci is constant. The foci F₁(4,-1) and F₂(4,7) share x-coordinate 4, making this a vertical hyperbola centered at (4,(-1+7)/2)=(4,3). The focal distance is 2c=8, so c=4, and |PF₂-PF₁|=2 gives 2a=2, so a=1. Using c²=a²+b² for hyperbolas, we get 16=1+b², so b²=15. The vertical hyperbola form is (y-k)²/a²-(x-h)²/b²=1, yielding (y-3)²/1-(x-4)²/15=1. A common error is using a²=4 thinking 2a=2 means a=2, but 2a represents the full difference, so a=1. Always verify that c>a, which confirms valid hyperbola geometry with our c=4>a=1.

To derive this ellipse equation, we start with the geometric definition: all points where the sum of distances to two foci is constant. The foci F₁(2,-1) and F₂(10,-1) have the same y-coordinate, so they're horizontal with center at ((2+10)/2, -1) = (6,-1). The distance between foci is 2c=8, so c=4, and PF₁+PF₂=14 gives 2a=14, so a=7. Using c²=a²-b², we get 16=49-b², so b²=33. For a horizontal ellipse centered at (h,k), the equation is (x-h)²/a²+(y-k)²/b²=1, giving (x-6)²/49+(y+1)²/33=1. Students often confuse which denominator is a² versus b², but a² is always under the term corresponding to the foci's direction (horizontal here). Verify by checking that the center-to-focus distance equals c=4.

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.

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.

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.

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.

To derive this ellipse equation, we start with the geometric definition: points where the sum of distances to foci equals a constant. The foci $F_1(1,1)$ and $F_2(1,9)$ share x-coordinate 1, indicating a vertical ellipse centered at $(1, \frac{1+9}{2}) = (1,5)$. The distance between foci is $2c=8$, so $c=4$, and $PF_1+PF_2=16$ gives $2a=16$, so $a=8$. Using $c^2=a^2-b^2$, we get $16=64-b^2$, so $b^2=48$. For a vertical ellipse centered at $(h,k)$, the form is $\frac{(y-k)^2}{a^2}+\frac{(x-h)^2}{b^2}=1$, giving $\frac{(y-5)^2}{64}+\frac{(x-1)^2}{48}=1$. Students often confuse which term gets $a^2$ versus $b^2$, but for vertical foci, $a^2$ goes under the y-term. Verify by checking that the center-to-focus distance equals $c=4$ and that $a>c$ for ellipse validity.

This problem requires deriving a hyperbola equation from its focus definition. A hyperbola is the set of all points P(x,y) where the absolute difference of distances to two foci is constant. The foci F₁(0,-5) and F₂(0,5) lie on the y-axis, making this a vertical hyperbola centered at the origin with c = 5. The condition |PF₂ - PF₁| = 6 means 2a = 6, so a = 3. For a hyperbola, c² = a² + b², so b² = c² - a² = 25 - 9 = 16, giving y²/9 - x²/16 = 1. A common mistake is using the ellipse relationship b² = a² - c² instead of the hyperbola relationship. Always verify that c > a for hyperbolas and use the correct formula.