If the vertex of the parabola is the point (-3, 0) and the directrix is the line x + 5 = 0, then its equation

NCERT Exemplar · Q39

(-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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The parabola opens to the right because the directrix is vertical and lies to the left of the vertex. The distance from vertex to directrix is $2$ , so $a = 2$ and the standard form is $y^2 = 4a(x - h)$ with $(h,k) = (-3,0)$ , giving $y^2 = 8(x + 3)$ . The correct option is (A).

We are given the vertex $(-3,0)$ and the directrix $x + 5 = 0$ , i.e., $x = -5$ . The vertex is the midpoint between the focus and the directrix, and the parabola is the set of points equidistant from the focus and the directrix. Since the directrix is a vertical line, the axis of the parabola is horizontal — the parabola opens either left or right.

Key idea: For a parabola with a horizontal axis, the standard form with vertex at $(h,k)$ is

(y - k)^2 = 4a(x - h)

where $a$ is the signed distance from the vertex to the focus (and also from the vertex to the directrix, but opposite direction). The sign of $a$ tells us the opening direction: if $a > 0$ , it opens to the right; if $a < 0$ , it opens to the left.

Now let’s work through it step by step.

Identify the vertex and directrix.

Vertex: $V = (-3, 0)$ .

Directrix: $x = -5$ .

Since the directrix is vertical, the axis is horizontal. The vertex lies to the right of the directrix (because $-3 > -5$ ), so the parabola opens to the right. Therefore $a$ will be positive.
Find the distance from the vertex to the directrix.

The distance along the horizontal axis is:

|(-3) - (-5)| = |2| = 2.

This distance is $|a|$ . Since the parabola opens right, $a = +2$ .

Write the standard equation. …

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Q.If the vertex of the parabola is the point (−3,0)(-3, 0)(−3,0) and the directrix is the line x+5=0x + 5 = 0x+5=0, then its equation is (A) y2=8(x+3)y^2 = 8(x + 3)y2=8(x+3) (B) x2=8(y+3)x^2 = 8(y + 3)x2=8(y+3) (C) y2=−8(x+3)y^2 = -8(x + 3)y2=−8(x+3) (D) y2=8(x+5)y^2 = 8(x + 5)y2=8(x+5)

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Q.If the vertex of the parabola is the point $(-3, 0)$ and the directrix is the line $x + 5 = 0$ , then its equation is
(A) $y^2 = 8(x + 3)$
(B) $x^2 = 8(y + 3)$
(C) $y^2 = -8(x + 3)$
(D) $y^2 = 8(x + 5)$

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