- Read the focus location and directrix equation from the graph. - On the y-axis, compare the vertical distance to the focus with the perpendicular distance to the directrix. - The required point lies halfway between the focus and the directrix along the axis perpendicular to the directrix.

1. From the graph, the focus is \(F=(0, 2)\), and the directrix is \(y=-2\). 2. Along the y-axis, the point equidistant from the focus and the directrix lies halfway between \(y=2\) and \(y=-2\). 3. That halfway value is \(y=0\), so \(V=(0, 0)\).

\(V=(0, 0)\)

- The axis of a parabola is perpendicular to its directrix. - The parabola opens toward its focus and away from its directrix. - Use the focus coordinate that stays constant along the axis of symmetry.

1. From the graph, the focus is \(F=(2, 1)\) and the directrix is \(x=-2\). 2. The directrix is vertical, so the parabola has a horizontal axis of symmetry. 3. The focus lies to the right of the directrix, so the parabola opens right. 4. The horizontal axis through the focus is \(y=1\).

The parabola opens right, and its axis of symmetry is \(y=1\).

- Start by finding the distance from each point to \(F\). - What direction gives the shortest path to the line \(y=0\)?

1. Point \(P\) is \(2\) units from \(F\) and \(2\) units from \(y=0\), so it lies on the parabola. 2. Point \(Q\) is \(1\) unit from \(F\) but \(3\) units from \(y=0\), so it does not lie on the parabola.

\(P=(2, 2)\) lies on the parabola: it is \(2\) units from the focus and \(2\) from the directrix. \(Q=(2, 3)\) does not: its distances are \(1\) and \(3\) units.

- The axis of a focus/directrix parabola is perpendicular to the directrix. - The parabola opens toward the side containing its focus.

1. For parabola \(A\), the directrix is horizontal, so the axis through the focus is vertical. The focus lies above the directrix, so the parabola opens up. 2. For parabola \(B\), the directrix is vertical, so the axis through the focus is horizontal. The focus lies left of the directrix, so the parabola opens left.

a) Vertical axis; opens up. b) Horizontal axis; opens left.

- A point lies on a parabola when two particular distances are equal. - Compare the distance from \(P\) to the focus with the perpendicular distance from \(P\) to the directrix. - Because the directrix is horizontal, its perpendicular distance depends only on the y-coordinate.

1. From the graph, \(F=(0, 3)\), the directrix is \(y=-3\), and \(P=(6, 3)\). 2. The distance from \(P\) to the focus is \(6\) units. 3. The perpendicular distance from \(P\) to the directrix is also \(6\) units. 4. Since the two distances are equal, \(P\) lies on the parabola.

Yes. \(P\) is \(6\) units from the focus and \(6\) units from the directrix, so it satisfies the parabola definition.

- Compare the vertex and focus to determine the axis direction and the signed focal distance. - The directrix is the same distance from the vertex as the focus, on the opposite side. - Use the standard form for a parabola with a vertical axis after identifying the vertex and focal distance.

1. The focus is \(3\) units above the vertex, so \(p=3\) and the parabola opens upward. 2. The directrix is the horizontal line \(3\) units below the vertex: \(y=-5\). 3. Using \((x-h)^2=4p(y-k)\) with \((h, k)=(1, -2)\) gives \((x-1)^2=12(y+2)\).

Directrix: \(y=-5\) Equation: \((x-1)^2=12(y+2)\)

- What is the shortest path from a point to a vertical line? - The focus is not directly above or beside \(P\). How can you find their distance? - What must be equal for a point to belong to the parabola?

1. The focus is \(F=(1, 1)\), the directrix is \(d: x=10\), and \(P=(5, 4)\). 2. The distance from \(P\) to \(F\) is \(\sqrt{(5-1)^2+(4-1)^2}=\sqrt{25}=5\) units. 3. Its shortest distance to \(d\) is the horizontal distance \(|10-5|=5\) units. The distances match, so \(P\) lies on the parabola.

Yes. Both distances are \(5\) units.

- Which line through the focus is perpendicular to the directrix? - Where is the point halfway between the focus and directrix along that axis? - What sign should the vertex-to-focus change have when the focus is below the vertex?

1. The axis is the vertical line \(x=2\), perpendicular to the horizontal directrix. 2. The vertex lies halfway between the focus's y-coordinate \(-4\) and directrix height \(2\), giving \(y=\frac{-4+2}{2}=-1\). Thus the vertex is \((2, -1)\). 3. From the vertex to the focus is a vertical change \(-4-(-1)=-3\), so \(p=-3\). The parabola opens down.

Vertex \((2, -1)\); \(p=-3\); opens down.

- Which variable is squared in the equation? - Match the equation to the appropriate vertex form rather than expanding it. - What does the sign of \(p\) tell you about the focus's side of the vertex?

1. Compare with the horizontal-parabola form \((y-k)^2=4p(x-h)\). Here \((h, k)=(-2, 1)\) and \(4p=12\), so \(p=3\). 2. The focus is \((h+p, k)=(1, 1)\), and the directrix is \(x=h-p=-5\). 3. Since \(p>0\), the parabola opens to the right.

Vertex \((-2, 1)\); focus \((1, 1)\); directrix \(x=-5\). It opens right because \(4p=12\) gives \(p=3>0\).

- Represent a general point on the parabola as \((x, y)\). - Write one distance to the focus and one perpendicular distance to the directrix, then set them equal. - Squaring both sides should allow several terms to cancel before you solve for the remaining relation.

1. From the graph, \(F=(0, 2)\) and the directrix is \(y=-2\). Let \(P=(x, y)\) be any point on the parabola. 2. The distance from \(P\) to the focus is \(\sqrt{x^2+(y-2)^2}\), and the perpendicular distance from \(P\) to the directrix is \(|y+2|\). 3. Set the squared distances equal: \(x^2+(y-2)^2=(y+2)^2\). 4. Expanding and simplifying gives \(x^2=8y\), so \(y=\frac{x^2}{8}\).

\(x^2=8y\), equivalently \(y=\frac{x^2}{8}\)

- Read the focus coordinates and directrix equation from the graph. - Set the distance from a general point \((x, y)\) to the focus equal to its perpendicular distance to the directrix. - The vertex lies halfway between the focus and directrix along the axis perpendicular to the directrix.

1. From the graph, \(F=(3, 2)\) and the directrix is \(y=-4\). Let \(P=(x, y)\) be a point on the parabola. 2. The distance to the focus is \(\sqrt{(x-3)^2+(y-2)^2}\), and the perpendicular distance to the directrix is \(|y+4|\). 3. Set the squared distances equal: \((x-3)^2+(y-2)^2=(y+4)^2\). 4. Expanding and simplifying gives \((x-3)^2=12(y+1)\). 5. The vertex lies halfway between the focus and directrix along the vertical axis \(x=3\), so the vertex is \((3, -1)\).

\((x-3)^2=12(y+1)\), with vertex \((3, -1)\)

- For a vertical directrix, the point-to-line distance is a horizontal distance. - Set the focus distance equal to the perpendicular distance to the directrix before simplifying. - After obtaining standard form, use the sign of the coefficient multiplying \((x-h)\) to determine the opening direction.

1. From the graph, \(F=(-1, 2)\) and the directrix is \(x=3\). Let \(P=(x, y)\) be a point on the parabola. 2. The distance to the focus is \(\sqrt{(x+1)^2+(y-2)^2}\), and the perpendicular distance to the directrix is \(|x-3|\). 3. Set the squared distances equal: \((x+1)^2+(y-2)^2=(x-3)^2\). 4. Expanding and simplifying gives \((y-2)^2=-8(x-1)\). 5. The negative coefficient on \((x-1)\) means the parabola opens left, toward the focus and away from the directrix.

\((y-2)^2=-8(x-1)\); the parabola opens left.

- Compare the equation with the standard form for a horizontal parabola. - The value multiplying \((x-h)\) equals \(4p\), where \(p\) is the directed vertex-to-focus distance. - The focus and directrix lie the same distance from the vertex on opposite sides along the axis of symmetry.

1. Compare \((y+1)^2=12(x-2)\) with \((y-k)^2=4p(x-h)\). This gives \(h=2\), \(k=-1\), and \(4p=12\), so \(p=3\). 2. The vertex is \((h, k)=(2, -1)\). 3. The focus is \((h+p, k)=(5, -1)\), and the directrix is \(x=h-p=-1\). 4. The vertex is \(3\) units from the focus. Its perpendicular distance to the line \(x=-1\) is also \(|2-(-1)|=3\) units, so the distances are equal.

Vertex: \((2, -1)\); focus: \((5, -1)\); directrix: \(x=-1\). The vertex is \(3\) units from both the focus and the directrix.

- Write one expression for the distance from \(P\) to the focus. - Write a second expression for the perpendicular distance from \(P\) to the horizontal directrix. - Use the defining equal-distance condition before simplifying the equation.

1. The distance from \(P=(4, y)\) to the focus is \(\sqrt{4^2+(y-2)^2}\). 2. The perpendicular distance from \(P\) to the directrix is \(|y+2|\). 3. Set the distances equal and square: \(16+(y-2)^2=(y+2)^2\). 4. Simplifying gives \(16-4y=4y\), so \(y=2\).

\(y=2\)

- Locate the vertex halfway between the focus and the directrix along the line perpendicular to the directrix. - Determine whether the focal parameter should be positive or negative from the opening direction. - For each proposed equation, compare both its vertex and its opening direction with the given focus-directrix data.

1. The vertex lies halfway between the focus and directrix along the horizontal axis, so the vertex is \((-1, 2)\). 2. The focus is \(2\) units left of the vertex, so \(p=-2\). 3. In \((y-k)^2=4p(x-h)\), this gives \((y-2)^2=-8(x+1)\), so Student A is correct. 4. Student B used the wrong sign for \(p\), making the parabola open right instead of toward the focus on the left. 5. Student C used the wrong horizontal shift; \((x-1)\) would place the vertex at \((1, 2)\) instead of halfway between the focus and directrix.

Student A is correct. Student B has the wrong sign, so the parabola opens in the wrong direction. Student C has the wrong horizontal shift, so its vertex is incorrect.

- Distance to a line is measured along a perpendicular, not toward an arbitrary point on that line. - Express the two distances from a general point \((x,y)\) before setting them equal. - After finding the equation, substitute the given x-coordinate and check both possible y-values.

1. A point \((x,y)\) is \(\sqrt{(x+2)^2+y^2}\) units from the focus and \(|x-4|\) units from the directrix. 2. Equating squared distances gives \((x+2)^2+y^2=(x-4)^2\). Simplification yields \(y^2=-12(x-1)\), a left-opening parabola. 3. At \(x=-2\), \(y^2=36\), so \(y=6\) or \(y=-6\). 4. Each of \((-2, 6)\) and \((-2, -6)\) is \(6\) units from the focus and \(6\) units from the directrix.

a) \(y^2=-12(x-1)\) b) \((-2, 6)\) and \((-2, -6)\), each \(6\) units from both the focus and directrix