Derive the Equation of a Parabola - Geometry

1. Identify the vertex for the parabola $y^2=-24x$.

   $(0,0)$. Standard form $y^2=-24x$ has vertex at origin.

2. Derive the equation from focus $(2,0)$ and directrix $x=-2$ using $PF=\text{dist}(P,d)$.

   $y^2=8x$. Setting $PF=\text{dist}(P,d)$ and squaring gives vertex at origin with $4p=8$.

3. What is the axis of symmetry for $(x-h)^2=4p(y-k)$?

   $x=h$. Vertical line through vertex for vertical-axis parabola.

4. Find the focus for the parabola $(x+1)^2=-8(y-3)$.

   $(-1,1)$. From $(x+1)^2=-8(y-3)$, vertex $(3)$ plus $p=-2$ gives $1$.

5. Derive the equation for focus $(0,0)$ and directrix $y=4$.

   $x^2=-8(y-2)$. Vertex at $(0,2)$ with $p=-2$, giving $4p=-8$.

6. What is the vertex of a parabola defined by focus $F$ and directrix $d$?

   Midpoint between $F$ and its perpendicular projection on $d$. Vertex lies on the perpendicular from focus to directrix, halfway between.

7. Derive the equation for focus $ (1,2) $ and directrix $ y = -2 $.

   $ $(x-1)^2$ = 8(y-0) $. Vertex at $ (1,0) $ with $ p=2 $, giving $ 4p=8 $.

8. Identify the directrix for the parabola $y^2=-24x$.

   $x=6$. From $y^2=-24x$, directrix is $|p|=6$ units right of vertex.

9. Find the directrix for the parabola $(y-4)^2=20(x-1)$.

   $x=-4$. From $(y-4)^2=20(x-1)$, vertex $(1)$ minus $p=5$ gives $-4$.

10. Derive the equation for focus $(3,-2)$ and directrix $x=-1$.

    $(y+2)^2=8(x-1)$. Vertex at $(1,-2)$ with $p=2$, giving $4p=8$.

11. Derive the equation for focus $(-1,4)$ and directrix $y=0$.

    $(x+1)^2=8(y-2)$. Vertex at $(-1,2)$ with $p=2$, giving $4p=8$.

12. For $(x-h)^2=4p(y-k)$, which direction does the parabola open if $p>0$?

    Upward. Positive $p$ opens in positive $y$-direction.

13. Identify the axis of symmetry for focus $(3,-2)$ and directrix $x=-1$.

    $y=-2$. Horizontal line through focus for horizontal-axis parabola.

14. For $(y-k)^2=4p(x-h)$, which direction does the parabola open if $p<0$?

    To the left. Negative $p$ opens in negative $x$-direction.

15. What is the standard form of a vertical-axis parabola with vertex $(h,k)$ and parameter $p$?

    $(x-h)^2=4p(y-k)$. Parabola opens up/down with squared $x$-term on left side.

16. What is the latus rectum length for $(x-h)^2=4p(y-k)$ or $(y-k)^2=4p(x-h)$?

    $|4p|$. Width of parabola at focus level equals absolute value of $4p$.

17. Identify the vertex for the parabola $x^2=16y$.

    $(0,0)$. Standard form $x^2=16y$ has vertex at origin.

18. For $(x-h)^2=4p(y-k)$, which direction does the parabola open if $p<0$?

    Downward. Negative $p$ opens in negative $y$-direction.

19. What is the defining property of a parabola using a focus $F$ and directrix line $d$?

    $PF=\text{dist}(P,d)$ for all points $P$ on the parabola. Each point on the parabola is equidistant from focus and directrix.

20. What is the distance formula from $P(x,y)$ to a line $Ax+By+C=0$?

    $\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$. Standard point-to-line distance formula using coefficients and normal vector.

21. What is the distance formula from $P(x,y)$ to a horizontal line $y=k$?

    $|y-k|$. Absolute value of difference between $y$-coordinate and line's $y$-value.

22. For $(y-k)^2=4p(x-h)$, which direction does the parabola open if $p>0$?

    To the right. Positive $p$ opens in positive $x$-direction.

23. Derive the equation for focus $(2,5)$ and directrix $y=1$.

    $(x-2)^2=8(y-3)$. Vertex at $(2,3)$ with $p=2$, giving $4p=8$.

24. Derive the equation for focus $ (7,0) $ and directrix $ x=1 $.

    $y^2=12(x-4)$. Vertex at $ (4,0) $ with $ p=3 $, giving $ 4p=12 $.

25. Identify the focus for the parabola $x^2=16y$.

    $(0,4)$. From $x^2=16y$, we have $4p=16$, so $p=4$ and focus at $(0,4)$.

26. What equation results from the focus-directrix definition using squared distances?

    $PF^2=\text{dist}(P,d)^2$. Squaring both sides eliminates absolute values in distance equation.

27. Derive the equation for focus $(-3,4)$ and directrix $x=1$.

    $(y-4)^2=-16(x+1)$. Vertex at $(-1,4)$ with $p=-4$, giving $4p=-16$.

28. Identify the focus for the parabola $y^2=-24x$.

    $(-6,0)$. From $y^2=-24x$, we have $4p=-24$, so $p=-6$ and focus at $(-6,0)$.

29. Derive the equation for focus $(5,6)$ and directrix $y=2$.

    $(x-5)^2=8(y-4)$. Vertex at $(5,4)$ with $p=2$, giving $4p=8$.

30. Identify the directrix for the parabola $x^2=16y$.

    $y=-4$. From $x^2=16y$, directrix is $p=4$ units below vertex.

31. Find the directrix for the parabola $(x-2)^2=12(y+1)$.

    $y=-4$. From $(x-2)^2=12(y+1)$, vertex $(-1)$ minus $p=3$ gives $-4$.

32. Derive the equation for focus $(0,0)$ and directrix $x=4$.

    $y^2=-8(x-2)$. Vertex at $(2,0)$ with $p=-2$, giving $4p=-8$.

33. Find the equation of the parabola with focus $(4,2)$ and directrix $x=0$.

    $(y-2)^2=8(x-2)$. Vertex at $(2,2)$ with $p=2$, giving $4p=8$.

34. Find the equation of the parabola with focus $(0,-1)$ and directrix $y=3$.

    $x^2=-8(y-1)$. Vertex at $(0,1)$ with $p=-2$, giving $4p=-8$.

35. Derive the equation for focus $(7,0)$ and directrix $x=1$.

    $y^2=12(x-4)$. Vertex at $(4,0)$ with $p=3$, giving $4p=12$.

36. What is $p$ in the standard parabola forms in terms of vertex and focus?

    $p=\text{signed distance from vertex to focus}$. Parameter represents directed distance along axis from vertex to focus.

37. Derive the equation for focus $(0,3)$ and directrix $y=-3$.

    $x^2=12y$. Distance from focus to directrix is $6$, so $4p=12$.

38. Find the vertex of the parabola with focus $(0,4)$ and directrix $y=0$.

    $(0,2)$. Vertex is midpoint between focus and directrix intersection point.

39. What is the relationship between the parabola axis and the directrix?

    The axis is perpendicular to the directrix. Axis passes through focus and is normal to directrix line.

40. What is the standard form of a horizontal-axis parabola with vertex $(h,k)$ and parameter $p$?

    $(y-k)^2=4p(x-h)$. Parabola opens left/right with squared $y$-term on left side.

41. Derive the equation from focus $(0,2)$ and directrix $y=-2$ using $PF=\text{dist}(P,d)$.

    $x^2=8y$. Setting $PF=\text{dist}(P,d)$ and squaring gives vertex at origin with $4p=8$.

42. Derive the equation for focus $(1,2)$ and directrix $y=-2$.

    $(x-1)^2=8(y-0)$. Vertex at $(1,0)$ with $p=2$, giving $4p=8$.

43. What is the axis of symmetry for $ $(y-k)^2$=4p(x-h) $?

    $y=k$. Horizontal line through vertex for horizontal-axis parabola.

44. Find the vertex of the parabola with focus $(6,1)$ and directrix $x=2$.

    $(4,1)$. Vertex is midpoint between focus and perpendicular foot on directrix.

45. Find $p$ for a parabola with vertex $(0,2)$ and focus $(0,5)$.

    $p=3$. Distance from vertex to focus along positive $y$-direction.

46. For $(y-k)^2=4p(x-h)$, what are the focus and directrix?

    Focus $(h+p,k)$; directrix $x=h-p$. Focus moves $p$ units right of vertex; directrix moves $p$ units left.

47. What is the distance formula from $P(x,y)$ to a vertical line $x=h$?

    $|x-h|$. Absolute value of difference between $x$-coordinate and line's $x$-value.

48. Derive the equation for focus $(0,-5)$ and directrix $y=1$.

    $x^2=-12(y+2)$. Vertex at $(0,-2)$ with $p=-3$, giving $4p=-12$.

49. Derive the equation for focus $(-2,0)$ and directrix $x=-6$.

    $y^2=8(x+4)$. Vertex at $(-4,0)$ with $p=2$, giving $4p=8$.

50. Derive the equation for focus $(2,-1)$ and directrix $x=6$.

    $(y+1)^2=-8(x-4)$. Vertex at $(4,-1)$ with $p=-2$, giving $4p=-8$.

51. Derive the equation for focus $ (0,-5) $ and directrix $ y=1 $.

    $x^2=-12(y+2)$. Vertex at $ (0,-2) $ with $ p=-3 $, giving $ 4p=-12 $.

52. Find the focus for the parabola $(y+2)^2=-12(x+5)$.

    $(-8,-2)$. From $(y+2)^2=-12(x+5)$, vertex $(-5)$ plus $p=-3$ gives $-8$.

53. Find $p$ for a parabola with vertex $(4,-1)$ and focus $(1,-1)$.

    $p=-3$. Distance from vertex to focus along negative $x$-direction.

54. Identify the axis of symmetry for focus $(2,5)$ and directrix $y=1$.

    $x=2$. Vertical line through focus for vertical-axis parabola.

55. What is the axis of symmetry for $(y-k)^2=4p(x-h)$?

    $y=k$. Horizontal line through vertex for horizontal-axis parabola.

56. What is the midpoint formula for points $A(x_1,y_1)$ and $B(x_2,y_2)$?

    $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$. Average of corresponding coordinates of the two endpoints.

57. Identify the equation of the parabola with focus $(p,0)$ and directrix $x=-p$.

    $y^2=4px$. Standard form with vertex at origin and horizontal orientation.

58. For $(x-h)^2=4p(y-k)$, what are the focus and directrix?

    Focus $(h,k+p)$; directrix $y=k-p$. Focus moves $p$ units above vertex; directrix moves $p$ units below.

59. Identify the equation of the parabola with focus $(0,p)$ and directrix $y=-p$.

    $x^2=4py$. Standard form with vertex at origin and $p$ from focus position.

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