Definition of a Parabola

A parabola is the locus of a point that moves in a plane such that its distance from a fixed point (the focus) is always equal to its perpendicular distance from a fixed line (the directrix). This is the key property of a parabola, which corresponds to an eccentricity of e = 1.

Standard Equation of a Parabola

Parabola

The simplest form of the equation is when the vertex is at the origin (0,0) and the axis of symmetry is one of the coordinate axes.

  • Right-handed Parabola: y2=4axy^2 = 4ax (opens to the right)

    • Vertex: (0, 0)
    • Focus: (a, 0)
    • Directrix: x=−ax = -a
    • Axis: y=0y = 0 (x-axis)

  • Left-handed Parabola: y2=−4axy^2 = -4ax (opens to the left, focus at (-a,0), directrix at x=a)

  • Upward Parabola: x2=4ayx^2 = 4ay (opens upwards, focus at (0,a), directrix at y=-a)

  • Downward Parabola: x2=−4ayx^2 = -4ay (opens downwards, focus at (0,-a), directrix at y=a)

Key Terminology

  • Axis of Symmetry: The line passing through the focus and perpendicular to the directrix. The parabola is perfectly symmetric about this line.
  • Vertex: The point where the parabola intersects its axis of symmetry. It is the turning point of the curve and lies exactly halfway between the focus and the directrix.
  • Focal Chord: Any chord (a line segment connecting two points on the parabola) that passes through the focus.
  • Latus Rectum: This is a special focal chord that is perpendicular to the axis of symmetry. Its length is a measure of the 'width' of the parabola at the focus.
    • For y2=4axy^2=4ax, the length of the latus rectum is 4a4a.
    • The endpoints of the latus rectum are (a, 2a) and (a, -2a).

Parametric Coordinates

For complex problems, describing points on a parabola with a single variable, or parameter, is incredibly useful. Any point on the parabola y2=4axy^2 = 4ax can be represented by the coordinates (at2,2at)(at^2, 2at). The parameter 't' is the slope of the line joining the origin to the point (x,y)(x,y) on the parabola if the vertex is at the origin.

Example 1: Finding Parameters from the Equation

Question: For the parabola y2=12xy^2 = 12x, find the coordinates of the focus, the equation of the directrix, and the length of the latus rectum.

Solution: Step 1: Compare the given equation with the standard form y2=4axy^2 = 4ax. 4a=12  ⟹  a=34a = 12 \implies a=3 Step 2: Use the value of 'a' to find the parameters.

  • Focus: (a,0)=(3,0)(a, 0) = \mathbf{(3, 0)}.
  • Directrix: x=−a  ⟹  x=−3x = -a \implies \mathbf{x = -3}.
  • Length of Latus Rectum: 4a=4(3)=124a = 4(3) = \mathbf{12}.

Example 2: Finding the Equation from Parameters

Question: Find the equation of the parabola with focus at (6,0) and directrix x=−6x=-6.

Solution: Step 1: Identify the type of parabola and the value of 'a'. The focus is at (a,0)(a,0) and the directrix is at x=−ax=-a. By comparison, we see that a=6\mathbf{a=6}. Since the focus is on the positive x-axis and the directrix is on the negative side, the parabola opens to the right. The vertex is at the origin (0,0). Step 2: Write the equation. The standard equation is y2=4axy^2=4ax. Substituting a=6a=6, we get: y2=4(6)x  ⟹  y2=24xy^2 = 4(6)x \implies \mathbf{y^2 = 24x}

Example 3: Parabola with a Shifted Vertex

Question: Find the equation of the parabola whose vertex is at (2,1) and focus is at (2,4).

Solution: Step 1: Analyze the vertex and focus. The vertex is (h,k)=(2,1)(h,k)=(2,1) and the focus is at (2,4)(2,4). The x-coordinates are the same, so the axis of symmetry is the vertical line x=2x=2. Since the focus is above the vertex, the parabola opens upwards. Step 2: Find the value of 'a'. The distance from the vertex to the focus is aa. a=(2−2)2+(4−1)2=3a = \sqrt{(2-2)^2 + (4-1)^2} = 3 Step 3: Write the equation. The equation of an upward-opening parabola with a shifted vertex is (x−h)2=4a(y−k)(x-h)^2 = 4a(y-k). Substituting the values, we get: (x−2)2=4(3)(y−1)  ⟹  (x−2)2=12(y−1)(x-2)^2 = 4(3)(y-1) \implies \mathbf{(x-2)^2 = 12(y-1)}

Example 4: Using the Latus Rectum

Question: Find the equation of the parabola whose latus rectum has endpoints (3,5) and (3,-3).

Solution: Step 1: Find the focus and the value of 'a'. The latus rectum is a vertical line segment (x=3x=3), so the parabola's axis is horizontal. The focus is the midpoint of the latus rectum: S=(3,5+(−3)2)=(3,1)S = \left(3, \frac{5+(-3)}{2}\right) = \mathbf{(3,1)} The length of the latus rectum is the distance between the endpoints: ∣5−(−3)∣=8|5 - (-3)| = 8. This length is equal to 4a4a. 4a=8  ⟹  a=24a=8 \implies \mathbf{a=2} Step 2: Determine the possible vertices and equations. The vertex is at a distance 'a' from the focus along the axis. Since the axis is horizontal (y=1), the vertex can be to the left or right of the focus.

  • Case 1 (Opens Right): Vertex is at (3−a,1)=(3−2,1)=(1,1)(3-a, 1) = (3-2, 1) = (1,1). The equation is (y−k)2=4a(x−h)  ⟹  (y−1)2=8(x−1)(y-k)^2=4a(x-h) \implies \mathbf{(y-1)^2 = 8(x-1)}.
  • Case 2 (Opens Left): Vertex is at (3+a,1)=(3+2,1)=(5,1)(3+a, 1) = (3+2, 1) = (5,1). The equation is (y−k)2=−4a(x−h)  ⟹  (y−1)2=−8(x−5)(y-k)^2=-4a(x-h) \implies \mathbf{(y-1)^2 = -8(x-5)}.