The equation of the parabola having focus at $(-1, -2)$ and the directrix $x - 2y + 3 = 0$ is ________.

The equation of the parabola having focus at (-1, – 1st PUC Mathematics

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(-1, -2)

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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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A point on the parabola is equidistant from the focus and the directrix. Setting $\text{(distance to focus)}=\text{(perp. distance to directrix)}$ and squaring gives $4x^2 + y^2 + 4xy + 4x + 32y + 16 = 0$ .

A parabola is the locus of points $P(x,y)$ equidistant from a fixed point (focus) and a fixed line (directrix). With focus $F(-1,-2)$ and directrix $x-2y+3=0$ :

1. Distance to the focus

d_F=\sqrt{(x+1)^2+(y+2)^2}

2. Perpendicular distance to the directrix (using $\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ with $a=1,\ b=-2,\ c=3$ )

d_D=\frac{|x-2y+3|}{\sqrt{5}}

3. Equate and square $d_F=d_D$ :

5\left[(x+1)^2+(y+2)^2\right]=(x-2y+3)^2

4. Expand each side

$\text{LHS}=5x^2+5y^2+10x+20y+25$ …

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

COMEDK 2025Set 2025-E1 markMCQ
$x^2=\lambda y$
›Reveal solutionSolution
For the parabola $x^2=\lambda y$ , the triangle formed by the vertex and the two ends of the latus rectum has area $\lambda^2/8$ . Setting this equal to 18 gives $\lambda = 12$ , option (B).

Concept & Intuition

Comparing $x^2=\lambda y$ with the standard form $x^2=4ay$ gives $4a=\lambda$ , so $a=\lambda/4$ . The focus is at $(0,a)$ and the latus rectum endpoints are $(\pm 2a, a)$ . The vertex is the origin.

Step-by-step solution

Latus rectum endpoints in terms of $\lambda$ .

$2a = \lambda/2$ and $a=\lambda/4$ , so the endpoints are $\left(-\dfrac{\lambda}{2}, \dfrac{\lambda}{4}\right)$ and $\left(\dfrac{\lambda}{2}, \dfrac{\lambda}{4}\right)$ .

Set up the triangle.

Vertex $(0,0)$ , base $=\lambda$ (distance between the endpoints), height $=\lambda/4$ (vertical distance from the vertex to the latus-rectum line). …
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COMEDK 2024Set 2024-E1 markMCQ
$y^2=4 a x$
›Reveal solutionSolution
The latus rectum length gives $a = 1.5$ , so the negative end is $(-1.5, 3)$ ; the chord from the vertex $(0,0)$ to that point has slope $-2$ , giving equation $2x + y = 0$ , which is option (B).

Concept & Intuition

The latus rectum of a parabola is the chord through the focus perpendicular to the axis. For $y^2 = 4ax$ , its endpoints are $(a, 2a)$ and $(a, -2a)$ . The "negative end" means the one with negative $x$ ? No — here "negative end" refers to the end with a negative $y$ -coordinate (since the vertex is at the origin and the latus rectum lies to the right). But wait: the problem says "passing through its vertex and the negative end of the latus rectum." The vertex is $(0,0)$ . The latus rectum endpoints are $(a, 2a)$ and $(a, -2a)$ . The "negative end" is $(a, -2a)$ . However, the chord goes from $(0,0)$ to $(a, -2a)$ . That gives slope $\frac{-2a}{a} = -2$ , so equation $y = -2x$ or $2x + y = 0$ . But we must first find $a$ from the given length.

Step-by-step

Find $a$ from the latus rectum length For $y^2 = 4ax$ , the length of the latus rectum is $4a$ . Given that this length is 6 units:

$4a = 6 \quad\Rightarrow\quad a = \frac{3}{2}.$

Identify the endpoints of the latus rectum

With $a = 1.5$ , the endpoints are $(1.5, 3)$ and $(1.5, -3)$ . The "negative end" means the one with the negative $y$ -coordinate: $(1.5, -3)$ .

Find the chord through the vertex and this point …
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COMEDK 2021Set 2021-B1 markMCQ
$x^2=-4y$
›Reveal solutionSolution
[!TLDR]

Focus (0,-2) with directrix y=2 gives a downward parabola $x^2=-8y$ .

Concept

For a vertical parabola with vertex at the origin, $x^2 = -4ay$ opens downward, with focus at $(0,-a)$ and directrix $y=a$ (CBSE/NCERT Class 11, Conic Sections).

Solution …
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KCET 2020Set A-11 markMCQ
$x^2 = 4ay$
›Reveal solutionSolution
The latus rectum length of a parabola $x^2 = 4ay$ is $4a$ . Substituting the given point $(2,1)$ into the equation gives $a = 1$ , so the latus rectum is $4$ .

The key here is to connect the given point to the parameter $a$ of the parabola. The standard form $x^2 = 4ay$ tells us two things: the parabola opens upward, and the length of the latus rectum (the chord through the focus perpendicular to the axis) is exactly $4a$ . So the problem reduces to finding $a$ from the condition that $(2,1)$ lies on the curve.

Substitute the point into the equation. Since $(2,1)$ satisfies $x^2 = 4ay$ , we plug $x = 2$ and $y = 1$ :

$2^2 = 4a \cdot 1 \quad \Rightarrow \quad 4 = 4a.$

Solve for $a$ .

Dividing both sides by $4$ gives $a = 1$ .

Recall the latus rectum length.

For any parabola $x^2 = 4ay$ , the latus rectum is the line segment through the focus $(0, a)$ parallel to the $x$ -axis, with endpoints at $x = \pm 2a$ . Its length is $4a$ .

With $a = 1$ , the length is $4 \times 1 = 4$ . …
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Q.The equation of the parabola having focus at (−1,−2)(-1, -2)(−1,−2) and the directrix x−2y+3=0x - 2y + 3 = 0x−2y+3=0 is ________.

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Q.The equation of the parabola having focus at $(-1, -2)$ and the directrix $x - 2y + 3 = 0$ is ________.

Unlock everything free for 14 days