Free math worksheets

- The vertex lies halfway between the focus and the directrix along the axis of symmetry. - The directrix is horizontal, so use the standard form for a vertical parabola. - The sign of the focal parameter determines whether the parabola opens upward or downward.

1. The vertex is halfway between the focus and directrix, so the vertex is \((0, 0)\). 2. The directed distance from the vertex to the focus is \(p=3\). 3. A vertical parabola with vertex \((h, k)\) has form \((x-h)^2=4p(y-k)\). 4. Substituting \(h=0\), \(k=0\), and \(p=3\) gives \(x^2=12y\). Since \(p>0\), the parabola opens upward.

The standard form is \(x^2=12y\). The vertex is \((0, 0)\), and the parabola opens upward.

- Match the equation to \((x-h)^2=4p(y-k)\). - The focus and directrix are each \(|p|\) units from the vertex. - The sign of \(p\) gives the opening direction.

1. Compare with \((x-h)^2=4p(y-k)\). Here \(h=-4\), \(k=1\), and \(4p=20\), so \(p=5\). 2. The vertex is \((-4, 1)\). 3. The focus is \((h, k+p)=(-4, 6)\). 4. The directrix is \(y=k-p=-4\). 5. The axis of symmetry is \(x=-4\), and the positive value of \(p\) means the parabola opens upward.

Vertex: \((-4, 1)\) Focus: \((-4, 6)\) Directrix: \(y=-4\) Axis of symmetry: \(x=-4\) Opening direction: upward

- Match the equation to \((y-k)^2=4p(x-h)\). - The focus and directrix are each \(|p|\) units from the vertex. - The sign of \(p\) gives the horizontal opening direction.

1. Compare with \((y-k)^2=4p(x-h)\). Here \(h=-1\), \(k=2\), and \(4p=24\), so \(p=6\). 2. The vertex is \((-1, 2)\). 3. The focus is \((h+p, k)=(5, 2)\). 4. The directrix is \(x=h-p=-7\). 5. The axis of symmetry is \(y=2\), and the positive value of \(p\) means the parabola opens to the right.

Vertex: \((-1, 2)\) Focus: \((5, 2)\) Directrix: \(x=-7\) Axis of symmetry: \(y=2\) Opening direction: right

- Locate the vertex halfway between the focus and the directrix. - Use the axis direction to choose the correct standard-form orientation. - The requested chord lies on the line through the focus perpendicular to the axis.

1. The vertex is midway between the focus and directrix, so it is \((1, 1)\). The focal parameter is \(p=3\). 2. The standard form is \((x-1)^2=12(y-1)\). 3. The requested chord is horizontal through the focus, so set \(y=4\). 4. Then \((x-1)^2=36\), giving \(x=-5\) or \(x=7\). 5. The chord endpoints are \((-5, 4)\) and \((7, 4)\).

a) \((x-1)^2=12(y-1)\) b) \((-5, 4)\) and \((7, 4)\)

- Transform the focus and vertex as individual points. - Transform the directrix by tracking its fixed coordinate. - Use the new vertex-to-focus distance in the appropriate standard form.

1. The original vertex is \((0, 0)\), halfway between the focus and directrix. 2. Transforming the focus gives \((2(1)-3, -2(0)+4)=(-1, 4)\). 3. Every point on the vertical directrix \(x=-1\) maps to the vertical line \(X=2(-1)-3=-5\). 4. The vertex maps to \((-3, 4)\). 5. The transformed focal distance is \(2\), and the parabola opens right, so its standard form is \((Y-4)^2=8(X+3)\).

Focus: \((-1, 4)\) Directrix: \(X=-5\) Vertex: \((-3, 4)\) Equation: \((Y-4)^2=8(X+3)\)

- The focus is centered on the latus rectum. - Use the latus rectum length to determine the focal parameter’s magnitude. - Use the opening direction to choose the sign of that parameter.

1. The focus is the midpoint of the latus rectum endpoints, so it is \((5, 3)\). 2. The latus rectum length is \(8\), which equals \(4|p|\). Thus \(|p|=2\). 3. Because the parabola opens left, \(p=-2\). 4. Since the focus is \((h+p, k)=(5, 3)\), the vertex is \((h, k)=(7, 3)\). 5. The directrix is \(x=h-p=9\). 6. The standard equation is \((y-3)^2=4p(x-7)=-8(x-7)\).

Focus: \((5, 3)\) Vertex: \((7, 3)\) Directrix: \(x=9\) Equation: \((y-3)^2=-8(x-7)\)

- A vertical directrix means the parabola has a horizontal axis. - Locate the vertex halfway between the focus and directrix. - Use the focus side of the vertex to determine the sign of the focal parameter.

1. The axis is horizontal because the directrix is vertical. 2. The vertex is halfway between the focus and directrix along \(y=-1\), so it is \((-1, -1)\). 3. The directed distance from the vertex to the focus is \(p=3\). 4. Use \((y-k)^2=4p(x-h)\) to obtain \((y+1)^2=12(x+1)\). 5. Since \(p>0\), the parabola opens to the right.

The standard form is \((y+1)^2=12(x+1)\). The vertex is \((-1, -1)\), and the parabola opens to the right.

- A vertical directrix gives a horizontal parabola. - Locate the vertex midway between the focus and directrix. - The focus lies on the side toward which the parabola opens.

1. The directrix is vertical, so the parabola has a horizontal axis. 2. The vertex is halfway between the focus and directrix, so it is \((1, 4)\). 3. The focus lies \(3\) units to the left of the vertex, so \(p=-3\). 4. Using \((y-k)^2=4p(x-h)\) gives \((y-4)^2=-12(x-1)\). 5. Since \(p<0\), the parabola opens to the left.

The standard form is \((y-4)^2=-12(x-1)\). The vertex is \((1, 4)\), and the parabola opens to the left.

- A horizontal directrix gives a vertical parabola. - Locate the vertex halfway between the focus and directrix. - The focus lies on the side toward which the parabola opens.

1. The directrix is horizontal, so the parabola has a vertical axis. 2. The vertex is halfway between the focus and directrix, so it is \((3, -2)\). 3. The focus lies \(3\) units below the vertex, so \(p=-3\). 4. Using \((x-h)^2=4p(y-k)\) gives \((x-3)^2=-12(y+2)\). 5. Since \(p<0\), the parabola opens downward.

The standard form is \((x-3)^2=-12(y+2)\). The vertex is \((3, -2)\), and the parabola opens downward.

- The focus and directrix lie on opposite sides of the vertex at equal distances. - A horizontal directrix gives a vertical parabola. - The sign of the focal parameter determines the opening direction.

1. The directrix is horizontal, so the parabola has a vertical axis through \(x=-2\). 2. The directrix is \(4\) units above the vertex, so the focus is \(4\) units below the vertex at \((-2, -1)\). 3. Thus the focal parameter is \(p=-4\). 4. Using \((x-h)^2=4p(y-k)\) gives \((x+2)^2=-16(y-3)\). 5. Since \(p<0\), the parabola opens downward.

Focus: \((-2, -1)\) Standard form: \((x+2)^2=-16(y-3)\) Opening direction: downward

- Locate the vertex midway between the focus and the nearest point on the directrix. - Use the signed vertex-to-focus distance as \(p\) in the vertical-parabola standard form. - The latus rectum is perpendicular to the axis and passes through the focus.

1. The axis is vertical, and the vertex is halfway between the focus and directrix, so the vertex is \((0,1)\). 2. The focus is \(4\) units above the vertex, so \(p=4\). 3. A vertical parabola with vertex \((h,k)\) has equation \((x-h)^2=4p(y-k)\). 4. Therefore, the equation is \(x^2=16(y-1)\). 5. The latus rectum passes through the focus, so set \(y=5\): \(x^2=16(4)=64\). 6. Thus, \(x=\pm8\), and the endpoints are \((-8,5)\) and \((8,5)\).

a) \(x^2=16(y-1)\) b) \((-8, 5)\) and \((8, 5)\)

- A horizontal directrix gives a vertical axis through the focus. - Locate the vertex halfway between the focus and the nearest point on the directrix. - Use the signed vertex-to-focus distance in \((x-h)^2=4p(y-k)\).

1. The directrix is horizontal, so the axis is vertical through the focus: \(x=3\). 2. The vertex is halfway between the focus and directrix, so it is \((3,1)\). 3. The focus is \(3\) units below the vertex, so \(p=-3\). 4. A vertical parabola has form \((x-h)^2=4p(y-k)\). 5. Therefore, \((x-3)^2=-12(y-1)\), and the negative value of \(p\) shows that it opens downward.

Equation: \((x-3)^2=-12(y-1)\) Vertex: \((3, 1)\) Axis of symmetry: \(x=3\) Opening direction: downward

- Start with the vertical-parabola standard form centered at the given vertex. - Substitute the known point to determine the signed focal parameter \(p\). - The focus is \(p\) units from the vertex along the axis, while the directrix is the same distance in the opposite direction.

1. A vertical parabola with vertex \((h,k)\) has form \((x-h)^2=4p(y-k)\). 2. Using \((h,k)=(-2,3)\), write \((x+2)^2=4p(y-3)\). 3. Substitute the point \((2,2)\): \(16=4p(-1)\), so \(p=-4\). 4. Therefore, the equation is \((x+2)^2=-16(y-3)\). 5. The focus is \((h,k+p)=(-2,-1)\). 6. The directrix is \(y=k-p=7\).

Equation: \((x+2)^2=-16(y-3)\) Focus: \((-2, -1)\) Directrix: \(y=7\)

- Start with horizontal-parabola standard form using the given vertex. - Substitute the known point to determine the signed focal parameter. - Move from the vertex by \(p\) toward the focus and by \(p\) in the opposite direction to locate the directrix.

1. A horizontal parabola with vertex \((h,k)\) has form \((y-k)^2=4p(x-h)\). 2. Substitute the vertex: \((y+2)^2=4p(x-1)\). 3. Use the point \((5,2)\): \(16=4p(4)\), so \(p=1\). 4. Therefore, the equation is \((y+2)^2=4(x-1)\). 5. The focus is \((h+p,k)=(2,-2)\). 6. The directrix is \(x=h-p=0\).

Equation: \((y+2)^2=4(x-1)\) Focus: \((2, -2)\) Directrix: \(x=0\)

- Locate the focus from the parabola's standard form. - First describe the locus of points equidistant from the focus and vertex. - Intersect that simpler locus with the parabola.

1. The vertex is \((0, 0)\), and \(4p=8\) gives focus \((2, 0)\). 2. Points equidistant from \((0, 0)\) and \((2, 0)\) lie on the perpendicular bisector \(x=1\). 3. Substituting \(x=1\) into \(y^2=8x\) gives \(y=\pm2\sqrt2\). 4. The distance from either point to the vertex is \(\sqrt{1+8}=3\), so the distance to the focus is also \(3\).

The points are \((1, 2\sqrt2)\) and \((1, -2\sqrt2)\). The common distance is \(3\).

- Locate the vertex halfway between the focus and directrix before identifying \(p\). - Substitute the fixed x-coordinate into the standard equation and include both square roots. - Compare the Euclidean focus distance with the horizontal distance to the directrix.

1. The vertex is midway between the focus and directrix, so it is \((0,0)\), and \(p=4\). 2. The standard equation is \(y^2=4px=16x\). 3. At \(x=5\), \(y^2=80\), so \(y=\pm4\sqrt{5}\). 4. The intersection points are \((5,4\sqrt{5})\) and \((5,-4\sqrt{5})\). 5. For either point, the distance to the directrix is \(5-(-4)=9\). 6. The distance to the focus is \(\sqrt{(5-4)^2+(4\sqrt{5})^2}=\sqrt{81}=9\), confirming equality.

a) \(y^2=16x\) b) \((5, 4\sqrt{5})\) and \((5, -4\sqrt{5})\) c) Each point is \(9\) units from both the focus and the directrix.

- Equate the distance from \((x, y)\) to the focus with its horizontal distance to the directrix. - Compare the simplified equation with \((y-k)^2=4p(x-h)\). - The latus rectum is perpendicular to the axis and passes through the focus.

1. For a point \((x, y)\) on the parabola, set the distance to the focus equal to the distance to the directrix: \(\sqrt{(x-3)^2+(y+2)^2}=|x+1|\). 2. Square both sides: \((x-3)^2+(y+2)^2=(x+1)^2\). 3. Simplifying gives \((y+2)^2=8(x-1)\). 4. Comparing with \((y-k)^2=4p(x-h)\) gives \(h=1\), \(k=-2\), and \(p=2\). 5. The vertex is \((1, -2)\), and the axis of symmetry is \(y=-2\). 6. The latus rectum passes through the focus, so set \(x=3\): \((y+2)^2=16\). 7. Thus \(y=2\) or \(y=-6\). The endpoints are \((3, 2)\) and \((3, -6)\), and the latus rectum has length \(8\).

a) \((y+2)^2=8(x-1)\) b) Vertex: \((1, -2)\); axis of symmetry: \(y=-2\) c) Endpoints: \((3, 2)\) and \((3, -6)\); length: \(8\)

- Interpret the square root as a point-to-point distance. - Compare \(x+2\) with the perpendicular distance to a vertical line. - After squaring, check the sign of the original right side on the entire resulting graph.

1. The left side is the distance from \((x, y)\) to \((2, 0)\). 2. When \(x\ge-2\), the expression \(x+2\) is the perpendicular distance to the line \(x=-2\). 3. Square the equation: \((x-2)^2+y^2=(x+2)^2\). 4. Simplifying gives \(y^2=8x\). 5. This is a parabola with vertex \((0, 0)\), focus \((2, 0)\), directrix \(x=-2\), and axis of symmetry \(y=0\). 6. Every point on \(y^2=8x\) has \(x\ge0\), so \(x+2>0\). Therefore every point on the parabola satisfies the original unsquared equation, and squaring adds no points.

Conic: parabola Standard equation: \(y^2=8x\) Vertex: \((0, 0)\) Focus: \((2, 0)\) Directrix: \(x=-2\) Axis of symmetry: \(y=0\) Squaring adds no points because every point on the parabola has \(x\ge0\), making the original right side positive.

- Compare the point's distance to the focus with its distance to a vertical line. - The opening direction determines which possible directrix is valid. - Once the focus and directrix are known, locate the point halfway between them along the axis.

1. The distance from \(P\) to the focus is \(\sqrt{(2-5)^2+(6-2)^2}=5\). 2. Let the vertical directrix be \(x=d\). The distance from \(P\) to the directrix is \(|2-d|\), so \(|2-d|=5\), giving \(d=-3\) or \(d=7\). 3. Because the parabola opens to the right, its directrix must lie to the left of the focus. Thus the directrix is \(x=-3\). 4. The vertex is midway between the focus and directrix, so it is \((1, 2)\). The focal parameter is \(p=4\), giving \((y-2)^2=16(x-1)\).

The directrix is \(x=-3\), and the parabola is \((y-2)^2=16(x-1)\).

- At a common point, use the circle radius as the distance to the focus. - Replace that focal distance with the equal distance to the directrix. - Use the resulting y-coordinate in the parabola equation.

1. A point on the parabola is equally distant from the focus and the directrix. 2. On the circle, the distance to the focus is \(6\). Therefore an intersection point must also be \(6\) units from the directrix \(y=-2\). 3. Since the parabola lies above the directrix, \(y+2=6\), so \(y=4\). 4. The parabola has vertex \((0, 0)\) and focal parameter \(p=2\), so its equation is \(x^2=8y\). 5. Substituting \(y=4\) gives \(x^2=32\), so \(x=\pm4\sqrt{2}\).

\((4\sqrt{2}, 4)\) and \((-4\sqrt{2}, 4)\)

- Derive the parabola’s standard form from the focus and directrix. - On a vertical line, the sign of the resulting squared quantity controls the number of intersections. - For two intersections, express the chord length as the difference of the two y-values.

1. The vertex is \((1, -1)\), and the focal parameter is \(p=2\), so the parabola is \((y+1)^2=8(x-1)\). 2. On \(x=t\), the intersections satisfy \((y+1)^2=8(t-1)\). 3. If \(t<1\), the right side is negative, so there are no intersections. If \(t=1\), there is one intersection. If \(t>1\), there are two intersections. 4. For \(t>1\), the y-values are \(-1\pm\sqrt{8(t-1)}\), so the chord length is \(2\sqrt{8(t-1)}\). 5. Setting this length equal to \(6\) gives \(8(t-1)=9\), so \(t=\frac{17}{8}\).

a) \(t<1\): zero points; \(t=1\): one point; \(t>1\): two points b) \(t=\frac{17}{8}\)

- Relate the chord line’s height to its distance from the directrix. - Find the two symmetric x-coordinates where that line meets the parabola. - The chord length is the separation of those two x-values.

1. The parabola is \(x^2=4py\). 2. If the horizontal chord line is \(y=t\), its distance from the directrix \(y=-p\) is \(d=t+p\), so \(t=d-p\). 3. At \(y=t\), the endpoints satisfy \(x=\pm2\sqrt{pt}\). 4. Therefore the chord length is \(4\sqrt{pt}=4\sqrt{p(d-p)}\). Real intersections require \(d\ge p\), and a nondegenerate chord requires \(d>p\). 5. For \(p=3\) and \(d=7\), the length is \(4\sqrt{3(4)}=8\sqrt{3}\).

General chord length: \(4\sqrt{p(d-p)}\); real intersections require \(d\ge p\), and a nondegenerate chord requires \(d>p\) When \(p=3\) and \(d=7\): \(8\sqrt{3}\)

- Use the equal-height points to locate the axis of symmetry. - Fit vertex form with the remaining point information. - Convert the fitted equation to focus-directrix standard form.

1. The points \((0, 5)\) and \((6, 5)\) are symmetric, so the axis is \(x=3\). 2. Write the parabola as \(y=a(x-3)^2+k\). 3. Substituting \((2, 1)\) gives \(a+k=1\). Substituting \((0, 5)\) gives \(9a+k=5\). 4. Subtracting gives \(a=\frac{1}{2}\), and then \(k=\frac{1}{2}\). 5. The equation is \((x-3)^2=2(y-\frac{1}{2})\), so \(4p=2\) and \(p=\frac{1}{2}\). 6. The vertex is \((3, \frac{1}{2})\), the focus is \((3, 1)\), and the directrix is \(y=0\).

Equation: \((x-3)^2=2(y-\frac{1}{2})\) Vertex: \((3, \frac{1}{2})\) Focus: \((3, 1)\) Directrix: \(y=0\)

- Express the circle’s radius in two different ways from the two geometric conditions. - The distance from a point to a vertical line is horizontal. - After equating the distances, simplify before identifying the locus.

1. Because the circle passes through \(F\), its radius is \(CF=\sqrt{(x-2)^2+y^2}\). 2. Because the circle is tangent to \(x=-2\), its radius is also the perpendicular distance from \(C\) to that line, \(x+2\). 3. Set the two radius expressions equal and square: \((x-2)^2+y^2=(x+2)^2\). 4. Expanding and simplifying gives \(y^2=8x\). 5. Comparing with \(y^2=4px\) gives \(p=2\). Therefore, the focus is \((2, 0)\), the directrix is \(x=-2\), and the vertex is \((0, 0)\).

Locus: \(y^2=8x\) Focus: \((2, 0)\) Directrix: \(x=-2\) Vertex: \((0, 0)\)

- First locate the vertex and latus rectum endpoints from the parabola's standard form. - Use symmetry to restrict the possible center of the circle. - Substitute the parabola equation into the circle to check for additional intersections.

1. The parabola has \(4p=8\), so \(p=2\). Its vertex is \((0, 0)\), and its latus rectum endpoints are \((2, 4)\) and \((2, -4)\). 2. By symmetry, the circle’s center has the form \((h, 0)\). 3. Since the circle passes through the origin, its radius is \(|h|\), so its equation is \((x-h)^2+y^2=h^2\). 4. Substituting \((2, 4)\) gives \((2-h)^2+16=h^2\), so \(h=5\). 5. The circle is \((x-5)^2+y^2=25\). 6. To find all common points, substitute \(y^2=8x\) into the circle: \((x-5)^2+8x=25\), which simplifies to \(x(x-2)=0\). 7. If \(x=0\), then \(y=0\). If \(x=2\), then \(y=\pm4\). Thus, there are no other intersections.

Circle: \((x-5)^2+y^2=25\) The only common points are \((0, 0)\), \((2, 4)\), and \((2, -4)\).

- Write the distance from a general point to the focus and its perpendicular distance to the vertical directrix. - Square the equal-distance equation and simplify it into horizontal-parabola standard form. - In part b), use the focus–directrix equality before substituting into the equation.

1. For a point \((x,y)\), the squared distance to the focus is \((x-4)^2+(y-1)^2\). 2. The squared distance to the directrix \(x=-2\) is \((x+2)^2\). 3. Equating these distances gives \((x-4)^2+(y-1)^2=(x+2)^2\). 4. Expanding and simplifying gives \((y-1)^2=12(x-1)\). 5. For a point on this parabola, the distance to the focus equals the distance to the directrix. A distance of \(5\) therefore gives \(x+2=5\), so \(x=3\). 6. Substitute \(x=3\) into the standard equation: \((y-1)^2=24\). 7. Thus, \(y=1\pm2\sqrt{6}\), giving the two required points.

a) \((y-1)^2=12(x-1)\) b) \((3, 1+2\sqrt{6})\) and \((3, 1-2\sqrt{6})\)

- Write the distance from a general point to the focus and the perpendicular distance to the vertical directrix. - Square the focus–directrix equality and simplify before identifying the vertex form. - For the y-axis intersections, substitute \(x=0\) into the standard equation.

1. For a point \((x,y)\), the squared distance to the focus is \((x+1)^2+(y-2)^2\). 2. The squared distance to the directrix is \((x-5)^2\). 3. Set the distances equal: \((x+1)^2+(y-2)^2=(x-5)^2\). 4. Expanding and simplifying gives \((y-2)^2=-12(x-2)\). 5. On the y-axis, \(x=0\), so \((y-2)^2=24\). 6. Therefore, \(y=2\pm2\sqrt{6}\), giving the two intersections.

a) \((y-2)^2=-12(x-2)\) b) \((0, 2+2\sqrt{6})\) and \((0, 2-2\sqrt{6})\)

- Parameterize the parabola point and locate its matching point on the directrix. - Form the two vectors with initial point \(P\). - Use a dot product to impose perpendicularity.

1. Parameterize \(P=(at^2, 2at)\). The focus is \(F=(a, 0)\), and the directrix is \(x=-a\). 2. The perpendicular foot is \(D=(-a, 2at)\). 3. The vectors from \(P\) are \(\overrightarrow{PF}=(a(1-t^2), -2at)\) and \(\overrightarrow{PD}=(-a(1+t^2), 0)\). 4. Their dot product is \(-a^2(1-t^2)(1+t^2)\). 5. Perpendicularity requires \(1-t^2=0\), so \(t=\pm1\). 6. The corresponding points are \((a, 2a)\) and \((a, -2a)\), the endpoints of the latus rectum.

\(P=(a, 2a)\) or \(P=(a, -2a)\)

- Translate the focus–directrix definition into an equality of squared distances. - After obtaining the parabola equation, substitute the line equation to find both intersections. - Use the distance formula only after identifying the two endpoint coordinates.

1. For a point \((x,y)\), the squared distance to the focus is \((x-2)^2+y^2\), and the squared distance to the directrix is \((x+2)^2\). 2. Equating the distances gives \((x-2)^2+y^2=(x+2)^2\). 3. Expanding and simplifying gives \(y^2=8x\). 4. Substitute \(y=x\): \(x^2=8x\), so \(x=0\) or \(x=8\). 5. The intersection points are \((0,0)\) and \((8,8)\). 6. Their distance is \(\sqrt{(8-0)^2+(8-0)^2}=8\sqrt{2}\).

a) \(y^2=8x\) b) \((0, 0)\) and \((8, 8)\) c) \(8\sqrt{2}\) units

- Derive both parabola equations from their common focus and separate directrices. - Subtract the equations to split the system into two simpler cases. - After finding both points, use the distance formula without rounding the radicals.

1. The parabola with directrix \(x=-4\) has equation \(y^2=8(x+2)\). 2. The parabola with directrix \(y=-4\) has equation \(x^2=8(y+2)\). 3. Subtracting the equations gives \((y-x)(x+y+8)=0\). 4. The case \(x+y=-8\) produces no real points satisfying both equations. 5. For \(y=x\), solve \(x^2=8(x+2)\) to obtain \(x=4\pm4\sqrt{2}\). 6. The intersection points are \((4-4\sqrt{2},4-4\sqrt{2})\) and \((4+4\sqrt{2},4+4\sqrt{2})\). 7. The change in each coordinate is \(8\sqrt{2}\), so the distance is \(\sqrt{(8\sqrt{2})^2+(8\sqrt{2})^2}=16\).

a) \((4-4\sqrt{2}, 4-4\sqrt{2})\) and \((4+4\sqrt{2}, 4+4\sqrt{2})\) b) \(16\) units

- Use the focus and directrix to write the parabola's equation. - Choose the directrix projection segment as a triangle base. - Replace the absolute \(x\)-coordinate with a nonnegative variable before solving.

1. The parabola is \(x^2=8y\), so \(y=\frac{x^2}{8}\). 2. Since \(D=(x, -2)\), the vertical segment \(PD\) has length \(y+2\). The perpendicular distance from \(F=(0, 2)\) to the line containing \(PD\) is \(|x|\). 3. Therefore, the triangle area is \(\frac12|x|(y+2)\). 4. Setting the area equal to \(8\) and substituting \(y=\frac{x^2}{8}\) gives \(\frac12|x|\left(\frac{x^2}{8}+2\right)=8\). 5. Let \(u=|x|\). Then \(u^3+16u-128=0\), which factors as \((u-4)(u^2+4u+32)=0\). 6. The only nonnegative solution is \(u=4\). Thus \(x=\pm4\) and \(y=2\).

The points are \((-4, 2)\) and \((4, 2)\).

- Set the distance from a general point to the focus equal to its perpendicular distance from the directrix. - Write the circle equation from its center and radius, then substitute the parabola equation. - Check each algebraic x-value against the requirement that \(y^2\) be nonnegative.

1. A point \((x, y)\) on the parabola is equally distant from the focus and directrix: \(\sqrt{(x-2)^2+y^2}=|x+2|\). 2. Square both sides and simplify: \((x-2)^2+y^2=(x+2)^2\), so \(y^2=8x\). 3. The vertex is halfway between the focus and directrix along the x-axis, so it is \((0, 0)\). 4. The circle centered at \((2, 0)\) with radius \(6\) has equation \((x-2)^2+y^2=36\). 5. Substitute \(y^2=8x\): \((x-2)^2+8x=36\). 6. Simplifying gives \(x^2+4x-32=0\), or \((x-4)(x+8)=0\). 7. The parabola \(y^2=8x\) requires \(x\ge0\), so \(x=-8\) is not valid. For \(x=4\), \(y^2=32\), so \(y=\pm4\sqrt2\).

a) \(y^2=8x\); vertex \((0, 0)\) b) \((4, 4\sqrt2)\) and \((4, -4\sqrt2)\)

- Equate the distance from \((x, y)\) to the focus with its vertical distance to the directrix. - Substitute the chord's y-coordinate into the standard parabola equation. - For the verification, compute one point-to-point distance and one perpendicular point-to-line distance.

1. For a point \((x, y)\) on the parabola, \(\sqrt{x^2+(y-4)^2}=|y+4|\). 2. Squaring and simplifying gives \(x^2+(y-4)^2=(y+4)^2\), so \(x^2=16y\). 3. The vertex is halfway between the focus and directrix, so it is \((0, 0)\). 4. On the line \(y=9\), the parabola equation gives \(x^2=144\), so \(x=\pm12\). 5. The chord endpoints are \((-12, 9)\) and \((12, 9)\), and the chord length is \(24\). 6. For \((12, 9)\), the distance to the focus is \(\sqrt{12^2+(9-4)^2}=\sqrt{169}=13\). 7. Its distance to the directrix \(y=-4\) is \(|9-(-4)|=13\), confirming the defining equality.

a) \(x^2=16y\); vertex \((0, 0)\) b) Endpoints: \((-12, 9)\) and \((12, 9)\); chord length: \(24\) c) At \((12, 9)\), both the focus distance and directrix distance equal \(13\).

- Represent the unknown focus with one coordinate on the stated axis. - Apply the equal-distance definition to the given point before writing the parabola equation. - The vertex is halfway between the focus and directrix along the axis.

1. Write the unknown focus as \(F=(0, p)\). 2. The point \(P=(6, 3)\) is \(6\) units from the directrix \(y=-3\). 3. By the focus-directrix definition, its distance to \(F\) is also \(6\): \(\sqrt{6^2+(3-p)^2}=6\). 4. Squaring gives \(36+(3-p)^2=36\), so \(p=3\). 5. Therefore the focus is \((0, 3)\). The vertex is halfway between the focus and directrix, so it is \((0, 0)\). 6. The focal parameter is \(3\), so the standard equation is \(x^2=4(3)y=12y\).

Focus: \((0, 3)\) Vertex: \((0, 0)\) Standard equation: \(x^2=12y\)

- Represent the parabola with an unknown vertex height and focal parameter. - Convert the chord's midpoint and length into its two endpoint coordinates. - Use the opening direction to choose the valid parameter value.

1. Write the parabola as \(x^2=4p(y-k)\), where \(p>0\). Its focus is \((0, k+p)\), so \(k+p=5\). 2. The chord has endpoints \((-4, 8)\) and \((4, 8)\). Substituting either endpoint gives \(16=4p(8-k)\). 3. Since \(k=5-p\), the equation becomes \(16=4p(3+p)\), or \(p^2+3p-4=0\). 4. The roots are \(p=1\) and \(p=-4\). The parabola opens upward, so \(p=1\). 5. Therefore, \(k=4\). The equation is \(x^2=4(y-4)\), the vertex is \((0, 4)\), and the directrix is \(y=k-p=3\).

Equation: \(x^2=4(y-4)\) Vertex: \((0, 4)\) Directrix: \(y=3\)

- Write an equation equating the distance to the focus with the distance to the line. - Locate the foot of the perpendicular from the focus to the directrix. - The vertex lies halfway between the focus and that perpendicular foot.

1. A point \((x, y)\) on the parabola has equal distances from \((2, 1)\) and the line \(x+y=0\). 2. Squaring the distances gives \((x-2)^2+(y-1)^2=\frac{(x+y)^2}{2}\). 3. Expanding and simplifying gives \(x^2-2xy+y^2-8x-4y+10=0\). 4. The perpendicular projection of \((2, 1)\) onto \(x+y=0\) is \((\frac{1}{2}, -\frac{1}{2})\). 5. The vertex is the midpoint of the focus and that projection, so it is \((\frac{5}{4}, \frac{1}{4})\). 6. The axis passes through the focus and is perpendicular to the directrix. Its equation is \(y=x-1\).

a) \(x^2-2xy+y^2-8x-4y+10=0\) b) Vertex: \((\frac{5}{4}, \frac{1}{4})\); axis: \(y=x-1\)

- Represent the unknown focus with one variable on the given axis. - Express the vertex and focal parameter in terms of that variable. - Substituting the given point leads to an equation that may have more than one valid root.

1. Let the focus be \((2, f)\). The vertex is halfway between the focus and directrix, so \(k=\frac{f-3}{2}\), and the focal parameter is \(p=\frac{f+3}{2}\). 2. The standard form is \((x-2)^2=4p(y-k)\). 3. Substituting \((6, 5)\) gives \(16=2(f+3)(5-\frac{f-3}{2})\), which simplifies to \(f^2-10f-23=0\). 4. Solving gives \(f=5+4\sqrt{3}\) or \(f=5-4\sqrt{3}\). Both values are greater than \(-3\), so both foci satisfy the stated condition. 5. For \(f=5+4\sqrt{3}\), \(k=1+2\sqrt{3}\) and \(4p=16+8\sqrt{3}\). 6. For \(f=5-4\sqrt{3}\), \(k=1-2\sqrt{3}\) and \(4p=16-8\sqrt{3}\).

Focus \((2, 5+4\sqrt{3})\): \((x-2)^2=(16+8\sqrt{3})(y-1-2\sqrt{3})\) Focus \((2, 5-4\sqrt{3})\): \((x-2)^2=(16-8\sqrt{3})(y-1+2\sqrt{3})\) Both parabolas have the required axis, directrix, and point, so the data allow two solutions.

- Express the chord endpoints from the parabola equation. - Use the focus-directrix property to find each equal side of the triangle. - Set the chord length equal to a focal side and check all resulting roots.

1. The parabola is \(x^2=8y\). At \(y=t\), the chord endpoints are \((\pm\sqrt{8t}, t)\), so \(t\ge0\). 2. The chord length is \(2\sqrt{8t}\). 3. Each endpoint lies on the parabola, so its distance to the focus equals its distance to the directrix. That common distance is \(t+2\). 4. The triangle is equilateral when \(2\sqrt{8t}=t+2\). 5. Squaring gives \(32t=t^2+4t+4\), or \(t^2-28t+4=0\). 6. Therefore \(t=14\pm8\sqrt{3}\). Both values are nonnegative and satisfy the unsquared equation.

\(t=14+8\sqrt{3}\) or \(t=14-8\sqrt{3}\)

- Use the point-to-line distance formula for the oblique directrix. - The vertex is halfway from the focus to its perpendicular projection on the directrix. - The axis passes through the focus in the direction normal to the directrix.

1. The squared distance from \((x,y)\) to the focus is \((x-2)^2+(y-1)^2\). 2. The squared perpendicular distance to the directrix is \(\frac{(2x-y-4)^2}{5}\). 3. Equating these distances gives \((x-2)^2+(y-1)^2=\frac{(2x-y-4)^2}{5}\). 4. Expanding and simplifying gives \(x^2+4xy+4y^2-4x-18y+9=0\). 5. The foot of the perpendicular from \(F\) to the directrix is \(D=\left(\frac{12}{5},\frac{4}{5}\right)\). 6. The vertex is the midpoint of \(F\) and \(D\), so it is \(\left(\frac{11}{5},\frac{9}{10}\right)\). 7. The axis passes through the focus and is perpendicular to the directrix. A direction vector is \((2,-1)\), so the axis equation is \(x+2y-4=0\).

a) Distance form: \((x-2)^2+(y-1)^2=\frac{(2x-y-4)^2}{5}\) Expanded form: \(x^2+4xy+4y^2-4x-18y+9=0\) b) Vertex: \(\left(\frac{11}{5}, \frac{9}{10}\right)\); axis: \(x+2y-4=0\)

- Represent the unknown vertical directrix by an equation \(x=d\). - Apply the equal-distance definition at the given point before locating a vertex. - Keep both solutions of the absolute-value equation and test each geometrically.

1. Write the vertical directrix as \(x=d\). 2. The distance from \(P=(2, 4)\) to the focus \((2, 0)\) is \(4\). 3. By the focus-directrix definition, the distance from \(P\) to the directrix must also be \(4\): \(|2-d|=4\). 4. Thus \(d=-2\) or \(d=6\). 5. If the directrix is \(x=-2\), the vertex is halfway between \(x=-2\) and the focus at \(x=2\), so the vertex is \((0, 0)\). Here \(p=2\), giving \(y^2=8x\). 6. If the directrix is \(x=6\), the vertex is \((4, 0)\). Here \(p=-2\), giving \(y^2=-8(x-4)\). 7. Both parabolas contain \((2, 4)\), so an opening direction or another independent condition is needed for uniqueness.

Possibility 1: directrix \(x=-2\), vertex \((0, 0)\), equation \(y^2=8x\) Possibility 2: directrix \(x=6\), vertex \((4, 0)\), equation \(y^2=-8(x-4)\) The data allow two vertical directrices at distance \(4\) from \(P\), so the parabola is not unique.