Length of latus rectum of parabola $y^2 = 4ax$ is (a) $4a$ (b) $2a$ (c) $8a$ (d) 0

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Length of latus rectum of parabola y² = 4ax is (a) 4a (b) 2a (c) 8a (d) 0

Question of 148

y^2 = 4ax

(a)

4a

(b)

2a

(c)

8a

(d) 0

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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, $a > 0$ .

y^2 = 4ax

Equation	Opens	Focus	Directrix
$y^2 = 4ax$	right	$(a, 0)$	$x = -a$
$y^2 = -4ax$	left	$(-a, 0)$	$x = a$
$x^2 = 4ay$	up	$(0, a)$	$y = -a$
$x^2 = -4ay$	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 $y^2 = 4ax$ 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:

\sqrt{(x - a)^2 + y^2} = x + a.

Squaring both sides:

(x - a)^2 + y^2 = (x + a)^2,

and expanding gives $y^2 = 4ax$ . 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 $y^2 = 12x$ , compare with $y^2 = 4ax$ : here $4a = 12$ , so $a = 3$ .

Vertex: $(0, 0)$
Focus: $(a, 0) = (3, 0)$
Directrix: $x = -3$ …

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Length of latus rectum of $y^2=4ax$ is $4a$ .

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 24 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$x^2 = -16y$
(a) $(0, -4)$
(b) $(0, 4)$
(c) $(4, 0)$
(d) $(-4, 0)$
›Reveal solutionSolution

A downward-opening parabola $x^2=-4ay$ has its focus at $(0,-a)$ ; match coefficients to find $a$ .

The standard form of a downward-opening parabola is $x^2 = -4ay$ , with focus at $(0,-a)$ and directrix $y=a$ .

…

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CBSE 2026Set ANNUAL1 mark
$y^2 = 4ax$
›Reveal solutionSolution

The focus of $y^2=4ax$ is $(a,0)$ .

The standard form $y^2=4ax$ represents a parabola opening rightward with vertex at the origin. Comparing with the general standard form, the focus is located on the positive $x$ - …

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CBSE 2026Set ANNUAL1 mark
$Q. (Same parabolic-reflector case study as 37(i).) Find the coordinates of the focus F.$
›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 $y^2=20x$ , matching the standard form $y^2=4ax$ with:

$4a = 20 \implies a = 5$ …

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CBSE 2026Set ANNUAL1 mark
$Q. (Same parabolic-reflector case study as 37(i).) What is the length of the latus rectum?$
›Reveal solutionSolution

For y² = 4ax, the latus rectum length is always 4a; with a = 5, this gives 20 cm.

For the parabola $y^2=4ax$ (here $a=5$ , from the same case study), the length of the latus rectum is:

$4a = 4(5) = 20$

…

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CBSE 2026Set 1A1 mark
$x^2 - 4y^2 = 4$
›Reveal solutionSolution

$x^2-4y^2=4$ becomes $\tfrac{x^2}{4}-\tfrac{y^2}{1}=1$ , so $a=2$ and transverse axis $=2a=4$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is an equation of a parabola?$
(a) $x^2 = 4ay$
(b) $x = 4ay$
(c) $4y + x = c$
(d) $x^2 + 4y^2 = c$
›Reveal solutionSolution

$x^2=4ay$ is the equation of a parabola.

…

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CBSE 2025Set ANNUAL1 markMCQ
$x^2 = 6y$
(a) $F\left(0, \dfrac{-3}{2}\right), y = \dfrac{3}{2}$
(b) $F\left(0, \dfrac{3}{2}\right), y = \dfrac{3}{2}$
(c) $F\left(0, \dfrac{3}{2}\right), y = \dfrac{-3}{2}$
(d) None of these
›Reveal solutionSolution

Matching $x^2=6y$ to the standard upward parabola $x^2=4ay$ 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 $x^2 = 4ay$ , with focus $(0,a)$ and directrix $y=-a$ .

Comparing $x^2 = 6y$ with $x^2 = 4ay$ :

$4a = 6 \implies a = \frac{3}{2}$ …

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CBSE 2025Set sz1 mark
$(0, -3)$
›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 $x^2=-12y$ .

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.

…

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CBSE 2024Set ANNUAL1 markMCQ
$x^2=5y$
(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 $x^2=4ay$ to identify $a$ , then use the standard directrix formula $y=-a$ .

The standard upward-opening parabola is $x^2 = 4ay$ , with directrix $y = -a$ .

Comparing $x^2 = 5y$ with $x^2 = 4ay$ :

$4a = 5 \implies a = \frac{5}{4}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$x^2 = -8y$
(a) $(2,0)$
(b) $(0,2)$
(c) $(-2,0)$
(d) $(0,-2)$
›Reveal solutionSolution

For $x^2=-8y$ , the focus is $(0,-2)$ .

The standard downward-opening parabola is $x^2=-4ay$ , with vertex at the origin and focus at $(0,-a)$ . Comparing $x^2=-8y$ with $x^2=-4ay$ gives $4a=8$ , so $a=2$ . T …

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CBSE 2024Set ANNUAL1 markMCQ
$y^2 = -4ax$
(a) $(-9, 0)$
(b) $(9, 0)$
(c) $(0, 0)$
(d) None of these.
›Reveal solutionSolution

The vertex of $y^2=-4ax$ is the origin $(0,0)$ .

The standard forms of a parabola with vertex at the origin and axis along the x-axis are $y^2=4ax$ (opens right) and $y^2=-4ax$ (opens left, for $a>0$ ). In both cases the vertex — the point where the para …

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CBSE 2024Set hz1 mark
$y^2 = -4x$
›Reveal solutionSolution

For $y^2 = -4ax$ (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 $y^2 = -4x$ . Compare this with the standard form $y^2 = -4ax$ , which opens to the left with vertex at the origin:

$4a = 4 \implies a = 1$

…

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Q.Length of latus rectum of parabola y2=4axy^2 = 4axy2=4ax is

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Q.Length of latus rectum of parabola $y^2 = 4ax$ is

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