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.

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.

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.

When you encounter an ellipse problem with given points and foci locations, you need to identify the semi-major axis $$a$$, semi-minor axis $$b$$, and then use the relationship $$c^2 = a^2 - b^2$$ to find the focal distance $$c$$.

Since the ellipse is centered at the origin with foci on the x-axis, it has the standard form $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, where $$a > b$$. The point $$(5, 0)$$ tells us the ellipse extends 5 units along the x-axis, so $$a = 5$$. The point $$(0, 3)$$ tells us it extends 3 units along the y-axis, so $$b = 3$$.

Now you can find the focal distance: $$c^2 = a^2 - b^2 = 25 - 9 = 16$$, so $$c = 4$$. The foci are located at $$(\pm c, 0) = (\pm 4, 0)$$, which is answer D.

Looking at the wrong answers: A gives $$(\pm 2, 0)$$, which would result from incorrectly calculating $$c^2 = 9 - 25 = -16$$ and taking $$c = 2$$. B shows $$(\pm \sqrt{16}, 0)$$, which equals $$(\pm 4, 0)$$ but leaves the square root unsimplified—this might tempt you if you forgot to simplify $$\sqrt{16} = 4$$. C gives $$(\pm \sqrt{34}, 0)$$, which comes from incorrectly adding $$a^2 + b^2 = 25 + 9 = 34$$ instead of subtracting.

Remember: for ellipses, always verify that $$a > b$$ when foci lie on the x-axis, and use $$c^2 = a^2 - b^2$$, not addition.

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.

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.

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.

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.

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 $$(y - k)^2 = 4p(x - h)$$ where $$(h,k)$$ is the vertex and $$p$$ is the distance from vertex to focus.

The vertex is at $$(-2, 4)$$, so $$h = -2$$ and $$k = 4$$. The focus is at $$(1, 4)$$. Since both points have the same y-coordinate and the focus is to the right of the vertex, this confirms rightward opening. The distance $$p$$ from vertex to focus is $$1 - (-2) = 3$$.

Substituting into the standard form: $$(y - 4)^2 = 4(3)(x - (-2))$$, which simplifies to $$(y - 4)^2 = 12(x + 2)$$.

Option A uses the wrong standard form $$(x + 2)^2 = 12(y - 4)$$, which describes a parabola opening upward, not rightward. Option B has $$(y - 4)^2 = 6(x + 2)$$, using the correct form but with $$4p = 6$$ instead of $$4p = 12$$. This represents a parabola with $$p = 1.5$$, placing the focus at $$(0.5, 4)$$ instead of $$(1, 4)$$. Option D also uses the correct orientation but has $$4p = 3$$, giving $$p = 0.75$$ and placing the focus at $$(-1.25, 4)$$.

Remember: for horizontal parabolas, use $$(y - k)^2 = 4p(x - h)$$, and always calculate $$p$$ as the actual distance between vertex and focus to get $$4p$$ correct.

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 $$(3, 1)$$ and the focus is at $$(3, -2)$$. Since both points have the same x-coordinate, this is a vertical parabola. The distance from vertex to focus is $$|1 - (-2)| = 3$$ units downward from the vertex.

The directrix must be 3 units upward from the vertex, placing it at $$y = 1 + 3 = 4$$. So the directrix is the horizontal line $$y = 4$$.

Now let's check which point lies on this line. Point D $$(0, 4)$$ has a y-coordinate of 4, so it lies on the directrix $$y = 4$$.

Why the other answers are wrong: Point A $$(3, -5)$$ lies on the vertical line through the vertex and focus, but below the focus. Point B $$(3, 1)$$ is the vertex itself, not on the directrix. Point C $$(3, -2)$$ 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.