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, – Class 11 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.

Showing the 12 most recent of 17 on this concept.

KEAM 2026Set eng-2026-04184 marksMCQ
$4y-x^{2}+4x-12=0$
›Reveal solutionSolution

In standard form $(x-2)^2=4(y-2)$ , the focus is one unit above the vertex $(2,2)$ , i.e. $(2,3)$ .

Rewrite $4y-x^{2}+4x-12=0$ :

$4y=x^{2}-4x+12=(x-2)^{2}+8\;\Rightarrow\;(x-2)^{2}=4(y-2).$ …
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KEAM 2026Set eng-2026-04194 marksMCQ
$y^{2} + 8x + 4y + 12 = 0$
›Reveal solutionSolution

Complete the square to standard form $(y+2)^2=-8(x+1)$ ; the latus rectum passes through the focus.

$y^2+8x+4y+12=0\Rightarrow (y+2)^2-4+8x+12=0\Rightarrow (y+2)^2=-8(x+1)$ . …
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KEAM 2026Set eng-2026-04204 marksMCQ
$225x^2+125y^2=28125$
›Reveal solutionSolution

Dividing by $28125$ gives $\frac{x^2}{125}+\frac{y^2}{225}=1$ with $a^2=225$ (major axis vertical), $b^2=125$ . Latus rectum $=\frac{2b^2}{a}=\frac{2\cdot125}{15}=\frac{50}{3}$ .

Divide $225x^2+125y^2=28125$ by $28125$ :

$\frac{x^2}{28125/225}+\frac{y^2}{28125/125}=1\;\Rightarrow\;\frac{x^2}{125}+\frac{y^2}{225}=1.$ …
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KEAM 2026Set eng-2026-04204 marksMCQ
$25x^2-36y^2-50x-72y-911=0$
›Reveal solutionSolution

Completing the square converts the hyperbola to $\frac{(x-1)^2}{36}-\frac{(y+1)^2}{25}=1$ with $a^2=36,\ b^2=25$ . Then $e=\sqrt{1+b^2/a^2}=\sqrt{61/36}=\frac{\sqrt{61}}{6}$ .

Group and complete squares in $25x^2-36y^2-50x-72y-911=0$ :
$25(x^2-2x)-36(y^2+2y)=911.$ $25[(x-1)^2-1]-36[(y+1)^2-1]=911,$ $25(x-1)^2-25-36(y+1)^2+36=911,$ $25(x-1)^2-36(y+1)^2=911+25-36=900.$
Divide by $900$ : …
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KEAM 2026Set eng-2026-04214 marksMCQ
$x-1=0$
›Reveal solutionSolution

Matching the directrix $x=1$ to the parabola gives $k^2+4k-32=0$ , so $k=4$ or $k=-8$ .

Rewrite $y^2-kx+8=0$ as $y^2=kx-8=k\left(x-\dfrac8k\right)$ , a parabola with vertex $\left(\dfrac8k,0\right)$ and $4p=k$ , so $p=\dfrac k4$ . The directrix is $x=\dfrac8k-\dfrac k4$ . Setting it to $1$ : …
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KEAM 2026Set eng-2026-04214 marksMCQ
$x^{2}=-9y$
›Reveal solutionSolution

For $x^2=4py$ the latus rectum length is $|4p|$ ; here $|{-9}|=9$ .

The parabola $x^2=-9y$ is of the form $x^2=4py$ with $4p=-9$ . The length of the latus …
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KEAM 2026Set eng-2026-04224 marksMCQ
$x$
›Reveal solutionSolution

Vertex is farther out than the focus, so the parabola opens toward the origin (leftward).

Vertex at $(6,0)$ , focus at $(3,0)$ . The focus lies to the left of the vertex, so the parabola opens in the $-x$ direction. …
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KEAM 2025Set eng-2025-04234 marksMCQ
$x^2 - 4x + 8y + 4 = 0$
›Reveal solutionSolution
The focus is $(2,-2)$ .

Concept and Intuition

Complete the square to put the parabola in vertex form $(x-h)^2=-4a(y-k)$ ; a downward-opening parabola has focus $a$ below the vertex.

Step-by-Step Solution

$x^2-4x+8y+4=0\Rightarrow (x-2)^2-4+8y+4=0\Rightarrow (x-2)^2=-8y$ .

Vertex $(2,0)$ ; $4a=8\Rightarrow a=2$ ; opens downward.

Focus $=(2,\,0-2)=(2,-2)$ . …
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KEAM 2025Set eng-2025-04264 marksMCQ
$y^{2} = 8x$
›Reveal solutionSolution

Writing $y^2=4ax$ with $4a=8$ gives $a=2$ , and the standard parametrization $(at^2,\,2at)=(2t^2,\,4t)$ .

Comparing $y^2=8x$ with $y^2=4ax$ gives $a=2$ . A general point of $y^2=4ax$ is $(at^2,\,2at)$ . With $a=2$ this is $(2t^2,\,4t)$ . …
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KEAM 2025Set eng-2025-04284 marksMCQ
$y^2 = 16x$
›Reveal solutionSolution

The two ends of the latus rectum of $y^2=16x$ are $(4,\pm8)$ ; the other end is $(4,-8)$ . …

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KEAM 2025Set eng-2025-04294 marksMCQ
$(x-1)^2 = 2(y-2)$
›Reveal solutionSolution

Comparing to $(x-h)^2=4p(y-k)$ : $h=1,k=2,4p=2\Rightarrow p=\frac12$ ; directrix is $y=k-p=\frac32$ , i.e. $2y-3=0$ . …

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KEAM 2024Set eng-2024-06054 marksMCQ
$y^2 = x$
›Reveal solutionSolution

Latus rectum of $y^2=4ax$ is $4a$ ; here $4a=1$ .

Write $y^2=x$ as $y^2=4ax$ with $4a=1$ , i.e. $a=\tfrac14$ . …
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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