Q.(Same parabolic-reflector case study as 37(i).) Find the coordinates of the focus F.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Parabola Standard Form
Standard Equations of a Parabola
A parabola is the set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix). To turn this definition into a clean equation, NCERT places the parabola in the simplest position: vertex at the origin with its axis along a coordinate axis. The equations you get are called the standard equations.
The four standard forms
Depending on which way the parabola opens, there are four standard equations. In each, a>0.
| Equation | Opens | Focus | Directrix |
|---|---|---|---|
| y2=4ax | right | (a,0) | x=−a |
| y2=−4ax | left | (−a,0) | x=a |
| x2=4ay | up | (0,a) | y=−a |
| x2=−4ay | down | (0,−a) | y=a |
For all four the vertex is at the origin (0,0) and the axis of the parabola is a coordinate axis.
Where y2=4ax comes from
Take the focus at F(a,0) and the directrix as the line x=−a. For a point P(x,y) on the parabola, its distance to the focus equals its distance to the directrix:
(x−a)2+y2=x+a.
Squaring both sides:
(x−a)2+y2=(x+a)2,
and expanding gives y2=4ax. The other three forms follow by turning the focus in a different direction.
Latus rectum
The latus rectum is the chord through the focus, perpendicular to the axis, with both ends on the parabola. For every standard parabola its length is 4a — the very same 4a that appears in the equation, which makes it quick to read off.
Worked example
For the parabola y2=12x, compare with y2=4ax: here 4a=12, so a=3.
- Vertex: (0,0)
- Focus: (a,0)=(3,0)
- Directrix: x=−3 …
From the case study's equation y2=20x=4ax, matching gives a=5; for a parabola in this standard form, the focus lies …
From the case study's equation y² = 20x = 4ax, the focus of a rightward-opening parabola is at (a, 0), giving a = 5.
From the same case study, the parabola's equation was found to be y2=20x, matching the standard form y2=4ax with:
4a=20⟹a=5 …
Showing the 12 most recent of 24 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The co-ordinates of the focus of the parabola x2=−16y are(a) (0,−4)(b) (0,4)(c) (4,0)(d) (−4,0)
›Reveal solutionSolution
A downward-opening parabola x2=−4ay has its focus at (0,−a); match coefficients to find a.
The standard form of a downward-opening parabola is x2=−4ay, with focus at (0,−a) and directrix y=a.
…
- CBSE 2026Set ANNUAL1 markQ.The coordinates of the focus of the parabola y2=4ax be ______.
›Reveal solutionSolution
The focus of y2=4ax is (a,0).
The standard form y2=4ax represents a parabola opening rightward with vertex at the origin. Comparing with the general standard form, the focus is located on the positive x- …
- CBSE 2026Set ANNUAL1 markQ.(Same parabolic-reflector case study as 37(i).) Find the coordinates of the focus F.
›Reveal solutionSolution
From the case study's equation y² = 20x = 4ax, the focus of a rightward-opening parabola is at (a, 0), giving a = 5.
From the same case study, the parabola's equation was found to be y2=20x, matching the standard form y2=4ax with:
4a=20⟹a=5 …
- CBSE 2026Set ANNUAL1 markQ.(Same parabolic-reflector case study as 37(i).) What is the length of the latus rectum?
›Reveal solutionSolution
For y² = 4ax, the latus rectum length is always 4a; with a = 5, this gives 20 cm.
For the parabola y2=4ax (here a=5, from the same case study), the length of the latus rectum is:
4a=4(5)=20
…
- CBSE 2026Set 1A1 markQ.Find the length of transverse axis of hyperbola x2−4y2=4.
›Reveal solutionSolution
x2−4y2=4 becomes 4x2−1y2=1, so a=2 and transverse axis =2a=4.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Which of the following is an equation of a parabola?(a) x2=4ay(b) x=4ay(c) 4y+x=c(d) x2+4y2=c
›Reveal solutionSolution
x2=4ay is the equation of a parabola.
…
- CBSE 2025Set ANNUAL1 markMCQQ.For the parabola x2=6y, the focus and the equation of directrix are respectively(a) F(0,2−3),y=23(b) F(0,23),y=23(c) F(0,23),y=2−3(d) None of these
›Reveal solutionSolution
Matching x2=6y to the standard upward parabola x2=4ay gives a=3/2, so focus (0,3/2) and directrix y=−3/2.
The standard form of an upward-opening parabola with vertex at the origin is x2=4ay, with focus (0,a) and directrix y=−a.
Comparing x2=6y with x2=4ay:
4a=6⟹a=23 …
- CBSE 2025Set sz1 markQ.The equation of the parabola with focus (0,−3) and directrix y=3 is ........................ .
›Reveal solutionSolution
Vertex at the origin (midpoint of focus (0,−3) and directrix y=3), opening downward since the focus is below the directrix, gives x2=−12y.
The vertex of a parabola is equidistant from the focus and the directrix, and lies on the axis joining them. Focus (0,−3) and directrix y=3 are both symmetric about y=0, so the vertex is at the origin (0,0), and the axis of the parabola is the y-axis.
…
- CBSE 2024Set ANNUAL1 markMCQQ.The equation of the directrix of the parabola x2=5y is(a) 4y+5=0(b) 2x+5=0(c) x+2y=0(d) None of these
›Reveal solutionSolution
Match the given parabola to the standard form x2=4ay to identify a, then use the standard directrix formula y=−a.
The standard upward-opening parabola is x2=4ay, with directrix y=−a.
Comparing x2=5y with x2=4ay:
4a=5⟹a=45
…
- CBSE 2024Set ANNUAL1 markMCQQ.Co-ordinates of the focus of the parabola x2=−8y is —(a) (2,0)(b) (0,2)(c) (−2,0)(d) (0,−2)
›Reveal solutionSolution
For x2=−8y, the focus is (0,−2).
The standard downward-opening parabola is x2=−4ay, with vertex at the origin and focus at (0,−a). Comparing x2=−8y with x2=−4ay gives 4a=8, so a=2. T …
- CBSE 2024Set ANNUAL1 markMCQQ.The vertex of the parabola y2=−4ax is:(a) (−9,0)(b) (9,0)(c) (0,0)(d) None of these.
›Reveal solutionSolution
The vertex of y2=−4ax is the origin (0,0).
The standard forms of a parabola with vertex at the origin and axis along the x-axis are y2=4ax (opens right) and y2=−4ax (opens left, for a>0). In both cases the vertex — the point where the para …
- CBSE 2024Set hz1 markQ.Directrix of parabola y2=−4x is .......
›Reveal solutionSolution
For y2=−4ax (a left-opening parabola), the directrix is the vertical line x=a; here a=1, so the directrix is x=1.
The given parabola is y2=−4x. Compare this with the standard form y2=−4ax, which opens to the left with vertex at the origin:
4a=4⟹a=1
…
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