IMOClass 11 › Parabola

Parabola

Standard Equations

A parabola is the set of all points in a plane that are equidistant from a fixed point and a fixed straight line. The fixed point is the focus and the fixed line is the directrix. This balance between a point and a line gives the parabola its single, sweeping branch.

Some vocabulary you will use throughout: the axis is the line through the focus perpendicular to the directrix (the curve's line of symmetry); the vertex is the point where the parabola meets its axis, sitting exactly midway between focus and directrix; and the latus rectum is the chord through the focus perpendicular to the axis. For all four standard forms below its length is the same:

$$\text{Length of latus rectum} = |4a|$$

Placing the vertex at the origin and the axis along a coordinate axis gives the four standard forms. The sign and which variable is squared tell you instantly which way the parabola opens:

EquationOpensFocusDirectrixAxis
$y^2 = 4ax$right$(a, 0)$$x = -a$$y = 0$
$y^2 = -4ax$left$(-a, 0)$$x = a$$y = 0$
$x^2 = 4ay$upward$(0, a)$$y = -a$$x = 0$
$x^2 = -4ay$downward$(0, -a)$$y = a$$x = 0$

In every case $a > 0$ is the distance from the vertex to the focus (and equally from the vertex to the directrix). A clean way to remember the orientation: if $y$ is squared the parabola opens horizontally, if $x$ is squared it opens vertically; the sign then chooses the direction.

Deeper Insight — why "equal distances" forces a curve, and where it shows up: The defining condition $PF = PM$ (distance to focus equals perpendicular distance to directrix) is deceptively simple, yet squaring it produces exactly a second-degree equation with only one squared variable — which is what makes a parabola distinct from a circle or an ellipse. The constant $4a$ is not decorative: it is forced by the algebra of equating those two distances, and it is the reason the latus rectum length is always $|4a|$. This focus-directrix balance is also why parabolas are everywhere in the physical world — the path of a projectile under gravity, the cross-section of a satellite dish, the reflector behind a torch bulb. A parabolic mirror gathers every ray parallel to the axis to the single focus, a property that falls straight out of the equal-distance definition. Master the four standard forms not as four separate facts but as one idea seen from four directions: square one variable, set it equal to $4a$ times the other, and let the sign point the way.

Parabola y squared = 4ax with focus, directrix, vertex and latus rectum Parabola: y² = 4ax x (axis) directrix x = −a vertex (0,0) focus (a, 0) latus rectum = 4a The four standard parabolas and their opening directions Four standard forms y²=4ax y²=−4ax x²=4ay x²=−4ay
Example 1: Find the focus, directrix and length of the latus rectum of $y^2 = 12x$.
  1. Compare with $y^2 = 4ax$: so $4a = 12$, giving $a = 3$.
  2. The parabola opens right, so focus is $(a, 0) = (3, 0)$.
  3. Directrix is $x = -a$, i.e. $x = -3$.
  4. Length of latus rectum $= 4a = 12$.

Answer: Focus $(3, 0)$, directrix $x = -3$, latus rectum $12$.

Example 2: Find the equation of the parabola with vertex at the origin, axis along the $y$-axis, passing through $(4, 2)$ and opening upward.
  1. Axis along $y$-axis and opening upward means the form $x^2 = 4ay$.
  2. Substitute the point $(4, 2)$: $4^2 = 4a(2) \Rightarrow 16 = 8a \Rightarrow a = 2$.
  3. So $4a = 8$ and the equation is $x^2 = 8y$.

Answer: $x^2 = 8y$.

Example 3: Find the equation of the parabola with focus $(0, -3)$ and directrix $y = 3$.
  1. The focus is below the origin and the directrix is above it, so the parabola opens downward: $x^2 = -4ay$.
  2. The focus is $(0, -a)$, so $-a = -3 \Rightarrow a = 3$.
  3. Thus $4a = 12$ and the equation is $x^2 = -12y$.

Answer: $x^2 = -12y$.

Example 4: For the parabola $y^2 = -8x$, state which way it opens and give its focus.
  1. Compare with $y^2 = -4ax$: so $4a = 8 \Rightarrow a = 2$.
  2. The negative sign with $y^2$ means it opens to the left.
  3. Focus of $y^2 = -4ax$ is $(-a, 0) = (-2, 0)$.

Answer: Opens left; focus $(-2, 0)$.

Example 5: Find the equation of the parabola with vertex at the origin and focus $(5, 0)$.
  1. The focus lies on the positive $x$-axis, so the parabola opens right: $y^2 = 4ax$.
  2. Focus is $(a, 0) = (5, 0)$, so $a = 5$.
  3. Then $4a = 20$, giving $y^2 = 20x$.

Answer: $y^2 = 20x$.

Example 6: Find the coordinates of the endpoints of the latus rectum of $y^2 = 16x$.
  1. Here $4a = 16 \Rightarrow a = 4$, so the focus is $(4, 0)$.
  2. The latus rectum is the vertical chord through the focus; substitute $x = 4$: $y^2 = 16(4) = 64 \Rightarrow y = \pm 8$.
  3. So the endpoints are $(4, 8)$ and $(4, -8)$ — a chord of length $16 = 4a$, as expected.

Answer: $(4, 8)$ and $(4, -8)$.

Quick recap
  • A parabola is the set of points equidistant from a focus and a directrix.
  • Four standard forms: $y^2 = \pm 4ax$ (opens horizontally), $x^2 = \pm 4ay$ (opens vertically).
  • For $y^2 = 4ax$: focus $(a,0)$, directrix $x = -a$, axis $y = 0$, vertex at origin.
  • The latus rectum has length $|4a|$ for all four forms.
  • $y$ squared → opens left/right; $x$ squared → opens up/down; the sign sets the direction.
✓ Quick check
The length of the latus rectum of x² = 24y is:
x² = 4ay gives a = 6. Length of latus rectum = 4a = 24.
The point on y² = 12x with y = -6 is:
36 = 12x gives x = 3.

Focus, Directrix and Latus Rectum

For y² = 4ax: focus (a, 0), directrix x = −a, axis the x-axis, and latus rectum 4a.

Example 1: Focus of y² = 16x.
4a = 16 → a = 4, focus (4, 0).
Example 2: Length of latus rectum of y² = 16x.
4a = 16.
Quick recap
  • y² = 4ax: focus (a, 0), directrix x = −a.
  • Latus rectum length = 4a.
✓ Quick check
Find the area of the triangle formed by joining the vertex of the parabola y² = 12x to the extremities of its latus rectum.
For y² = 12x, 4a = 12, a = 3. Vertex is (0,0). Extremities are (3, 6) and (3, -6). The triangle has base 12 and height 3. Area = 1/2 × 12 × 3 = 18.
If a parabola has its vertex at (3, 2) and focus at (5, 2), what is the equation of its directrix?
The distance from vertex to focus is a = 5 - 3 = 2. The directrix is perpendicular to the horizontal axis (y=2) and is 2 units to the left of the vertex: x = 3 - 2 = 1.

Reading the Elements

Reduce a given equation to a standard form, identify a, then read off vertex, focus, directrix and latus rectum.

Example 1: Vertex of x² = −4y.
The origin (0, 0).
Example 2: Directrix of x² = 8y.
4a = 8 → a = 2, directrix y = −2.
Quick recap
  • Match the equation to a standard form to find a.
  • Then read vertex, focus, directrix, latus rectum.
✓ Quick check
A point on y² = 16x has y-coordinate 8. The x-coordinate is:
64 = 16x gives x = 4.
A parabolic suspension bridge in Mumbai has a span of 100 m and a maximum dip of 20 m from the supports. Assuming the lowest point is the origin, find the equation of the cable.
Let the vertex be (0,0). The parabola opens upwards: x² = 4ay. It passes through (50, 20). 50² = 4a(20) → 2500 = 80a → a = 2500/80 = 125/4. Equation: x² = 4(125/4)y = 125y.
Ready to test this chapter?
Take the Chapter Test →