Find the equation of each of the following parabolas: (a) Directrix x = 0, focus at (6, 0); (b) Vertex at (0,

NCERT Exemplar · Q28

Q. Find the equation of each of the following parabolas:

(a) Directrix

x = 0

, focus at

(6, 0)

;

(b) Vertex at

(0, 4)

, focus at

(0, 2)

;

(-1, -2)

, directrix

x - 2y + 3 = 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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Using distance-to-focus $=$ distance-to-directrix: (a) $y^2 = 12(x-3)$ ;

(b) $x^2 = -8(y-4)$ ;

A parabola is the locus of points equidistant from the focus and the directrix. Apply this definition to each part.

(a) Directrix $x = 0$ , focus $(6, 0)$ .

\sqrt{(x-6)^2 + y^2} = |x|

Squaring: $(x-6)^2 + y^2 = x^2 \;\Rightarrow\; y^2 - 12x + 36 = 0$

y^2 = 12(x - 3)

(b) Vertex $(0, 4)$ , focus $(0, 2)$ .

The focus is below the vertex, so the parabola opens downward with $a = 2$ and directrix $y = 6$ .

\sqrt{x^2 + (y-2)^2} = |y - 6|

Squaring: $x^2 + y^2 - 4y + 4 = y^2 - 12y + 36 \;\Rightarrow\; x^2 + 8y - 32 = 0$

x^2 = -8(y - 4)

(c) Focus $(-1, -2)$ , directrix $x - 2y + 3 = 0$ . …

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Q.Find the equation of each of the following parabolas:

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