Introduction to Parabola

Q: What are the parametric equations of a parabola?

For y^2 = 4ax, every point can be written as (at^2, 2at) for some real t. The value t = 0 gives the vertex, positive t gives the upper half and negative t the lower half. The parameter simplifies chord, tangent, normal and focal chord equations.

Q: How do you find the focal distance of a point on a parabola?

For a point (x1, y1) on y^2 = 4ax, the focal distance equals x1 + a. In parametric form, if the point is (at^2, 2at), the focal distance is a(1 + t^2). For x^2 = 4ay the focal distance is y1 + a.

Q: What is the condition for a line to touch a parabola?

The line y = mx + c with m not zero is tangent to y^2 = 4ax if and only if c = a/m. Equivalently, the tangent with slope m is y = mx + a/m, touching the parabola at (a/m^2, 2a/m). A line with m = 0 meets the curve exactly once but is a secant, not a tangent.

Q: What is a focal chord and what is special about it?

A focal chord is any chord passing through the focus. If its endpoints have parameters t1 and t2, then t1 t2 = -1, so the two ends lie on opposite sides of the axis. Its length when it makes angle alpha with the axis is 4a cosec^2 alpha, and the semi-latus rectum 2a is the harmonic mean of its two segments.

Q: How do you determine whether a point lies inside or outside a parabola?

For y^2 = 4ax, evaluate S1 = y1^2 - 4ax1. If S1 is negative the point is inside, zero means on the parabola, positive means outside. For a shifted parabola use S1 = (y1 - k)^2 - 4a(x1 - h).

Q: Why is the sum of reciprocals of focal chord segments equal to 1/a?

For a focal chord PSQ of y^2 = 4ax, PS = a(1 + t^2) and SQ = a(1 + 1/t^2). Their reciprocals add to 1/a. Equivalently, the semi-latus rectum 2a is the harmonic mean of PS and SQ.

Q: How is a parabola different from an ellipse or a hyperbola?

The parabola has eccentricity exactly 1, one focus, one directrix, a single unbounded branch, no centre and no asymptote. An ellipse has e less than 1 and is closed with two foci; a hyperbola has e greater than 1 with two branches, two foci and two asymptotes. All parabolas are similar, differing only in the value of a.

Q: How can you tell whether a general second-degree equation represents a parabola?

Write it as Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0. It represents a parabola when H^2 = AB and the determinant of the matrix with rows (A, H, G), (H, B, F), (G, F, C) is non-zero. If H is also zero the axis is parallel to a coordinate axis and the equation reduces by completing the square. If the determinant is zero the equation degenerates into parallel lines.

MathsParabolaFor JEE aspirants

A parabola is the conic section with eccentricity $e = 1$ : the locus of a point that stays equidistant from a fixed point (the focus) and a fixed straight line (the directrix). That single balancing condition produces the four standard equations $y^{2} = \pm 4 a x$ and $x^{2} = \pm 4 a y$ , each with vertex at the origin, focus at distance $a$ from the vertex along the axis, and latus rectum of length $4 a$ . This concept builds the whole foundation: the standard and shifted forms, the parametric point $(a t^{2}, 2 a t)$ , the focal distance $x_{1} + a$ , the position test $S_{1} = y_{1}^{2} - 4 a x_{1}$ , how a line cuts the curve, and the focal-chord relation $t_{1} t_{2} = - 1$ on which every later tangent and normal result is built.

Key Formulas - Quick Reference

Definition: $S P = P M$ , i.e. distance from focus $=$ perpendicular distance from directrix. Eccentricity $e = 1$ .
Standard form (right-opening): $y^{2} = 4 a x$ with vertex $(0, 0)$ , focus $(a, 0)$ , directrix $x + a = 0$ , axis $y = 0$ , latus rectum $= 4 a$ .
Four orientations: $y^{2} = 4 a x$ (right), $y^{2} = - 4 a x$ (left), $x^{2} = 4 a y$ (up), $x^{2} = - 4 a y$ (down), with $a > 0$ .
Shifted form: $(y - k)^{2} = 4 a (x - h)$ has vertex $(h, k)$ , focus $(h + a, k)$ , directrix $x = h - a$ , axis $y = k$ .
Parametric form of $y^{2} = 4 a x$ : $x = a t^{2}, y = 2 a t$ ; any point is $P (t) = (a t^{2}, 2 a t)$ .
Focal distance: for $P (a t^{2}, 2 a t)$ on $y^{2} = 4 a x$ , $S P = x_{P} + a = a (1 + t^{2})$ .
Position of a point: $S_{1} = y_{1}^{2} - 4 a x_{1}$ ; inside if $S_{1} < 0$ , on if $S_{1} = 0$ , outside if $S_{1} > 0$ .
Line and parabola ( $m \neq = 0$ ): $y = m x + c$ meets $y^{2} = 4 a x$ in two points if $m c < a$ , one point if $m c = a$ , none if $m c > a$ .
Condition of tangency: $c = \frac{a}{m}$ , with point of contact $(\frac{a}{m ^{2}}, \frac{2 a}{m})$ .
Chord joining parameters $t_{1}, t_{2}$ : $2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0$ , of length $a ∣ t_{1} - t_{2} ∣ (t_{1} + t_{2})^{2} + 4$ .
Focal chord condition: $t_{1} t_{2} = - 1$ ; endpoints $(a t^{2}, 2 a t)$ and $(\frac{a}{t ^{2}}, - \frac{2 a}{t})$ .
Length of focal chord: $a (t + \frac{1}{t})^{2}$ , or $4 a csc^{2} α$ if it makes angle $α$ with the axis. Minimum value $4 a$ (the latus rectum).
Harmonic mean property: $\frac{1}{S P} + \frac{1}{S Q} = \frac{1}{a}$ for any focal chord $P S Q$ .

1. Definition and the Standard Equation

Definition. A parabola is the locus of a point that moves so that its distance from a fixed point

S

(the focus) is always equal to its perpendicular distance from a fixed straight line (the directrix), the focus not lying on the directrix. In conic language, it is the conic with eccentricity

e = 1

Figure 1: The defining property of a parabola. For every point

P

on the curve, the focal distance

S P

equals the perpendicular distance

P M

to the directrix, which is exactly what

e = 1

means.

Deriving $y^{2} = 4 a x$ from the definition

Nothing is lost by choosing convenient axes. Let the distance from the focus to the directrix be $2 a$ (with $a > 0$ ), put the origin at the midpoint of that gap, and take the $x$ -axis along the line through the focus perpendicular to the directrix. Then

S = (a, 0), directrix : x = - a .

For a point $P (x, y)$ on the locus, the condition $S P = P M$ gives

(x - a)^{2} + y^{2} = ∣ x + a ∣.

Squaring both sides,

x^{2} - 2 a x + a^{2} + y^{2} = x^{2} + 2 a x + a^{2} ⟹ y^{2} = 4 a x .

The origin satisfies the equation, so the curve passes through the midpoint of the focus and the directrix. That point is the vertex. Notice also that replacing $y$ by $- y$ leaves the equation unchanged, so the parabola is symmetric about the $x$ -axis; that line of symmetry is the axis of the parabola.

Why the latus rectum is $4 a$ : put

x = a

(the vertical line through the focus) into

y^{2} = 4 a x

. Then

y^{2} = 4 a^{2}

, so

y = \pm 2 a

. The chord runs from

(a, - 2 a)

(a, 2 a)

, so its length is

4 a

. The number

4 a

therefore controls the entire size of the parabola: a large

a

gives a wide, shallow curve, a small

a

a narrow, steep one.

2. Elements of the Standard Parabola

Figure 2: The standard right-opening parabola

y^{2} = 4 a x

. Vertex

A (0, 0)

, focus

S (a, 0)

, directrix

x + a = 0

, axis

y = 0

, and latus rectum

L L^{'}

of length

4 a

with

L (a, 2 a)

L^{'} (a, - 2 a)

Element	Value for $y^{2} = 4 a x$
Vertex $A$	$(0, 0)$
Focus $S$	$(a, 0)$
Directrix	$x + a = 0$ , a vertical line at distance $a$ to the left of the vertex
Axis	$y = 0$ (the $x$ -axis)
Latus rectum	The chord $x = a$ , of length $4 a$
Ends of latus rectum	$L (a, 2 a)$ and $L^{'} (a, - 2 a)$
Focal distance of $(x_{1}, y_{1})$	$x_{1} + a$
Eccentricity	$e = 1$

The four standard forms

With the vertex at the origin and the axis along a coordinate axis, a parabola takes one of four forms. Throughout, $a > 0$ ; the choice of which variable is squared decides whether the axis is horizontal or vertical, and the sign decides the direction of opening.

y^{2} = 4 a x, y^{2} = - 4 a x, x^{2} = 4 a y, x^{2} = - 4 a y .

Figure 3: The four standard orientations. In each panel the focus (solid dot) sits inside the curve and the directrix (dashed line) lies on the far side of the vertex, at the same distance

a

Equation	Opens	Focus	Directrix	Axis	Parametric point
$y^{2} = 4 a x$	Right	$(a, 0)$	$x = - a$	$y = 0$	$(a t^{2}, 2 a t)$
$y^{2} = - 4 a x$	Left	$(- a, 0)$	$x = a$	$y = 0$	$(- a t^{2}, 2 a t)$
$x^{2} = 4 a y$	Up	$(0, a)$	$y = - a$	$x = 0$	$(2 a t, a t^{2})$
$x^{2} = - 4 a y$	Down	$(0, - a)$	$y = a$	$x = 0$	$(2 a t, - a t^{2})$

In every case the vertex is $(0, 0)$ , the latus rectum has length $4 a$ , and the focus lies inside the curve while the directrix lies on the opposite side of the vertex. A quick memory hook: the parabola always opens towards its focus and runs away from its directrix.

Vocabulary you will need

Focal distance - the distance from a point on the parabola to the focus. For $y^{2} = 4 a x$ it equals $x_{1} + a$ ; for $x^{2} = 4 a y$ it equals $y_{1} + a$ .
Focal chord - any chord that passes through the focus.
Double ordinate - any chord perpendicular to the axis of symmetry. For the point at parameter $t$ , the double ordinate through it has length $4 a ∣ t ∣$ .
Latus rectum - the double ordinate that passes through the focus, i.e. the focal chord perpendicular to the axis. Its length is $4 a$ , and it is the shortest focal chord.
Semi-latus rectum - half the latus rectum, $2 a$ . It is the harmonic mean of the two parts of any focal chord.

Four quick facts:

Perpendicular distance from focus to directrix $= 2 a =$ half the latus rectum.
The vertex is the midpoint of the focus and the point where the directrix meets the axis.
Two parabolas are called equal if they have the same latus rectum (the same $a$ ), whatever their position or orientation. All parabolas are similar to one another, unlike ellipses, whose shape depends on $e$ .
A parabola has no centre and no asymptote, and it consists of one single unbounded branch.

Where this shows up in physics: because

S P = P M

for every point, a ray travelling parallel to the axis is reflected by a parabolic mirror straight through the focus. That is why satellite dishes, torch reflectors and car headlights are parabolic in cross section. The projectile path under uniform gravity is also a parabola.

3. Shifted Parabola: Vertex Not at the Origin

When the vertex moves to $(h, k)$ but the axis stays parallel to a coordinate axis, simply replace $x$ by $(x - h)$ and $y$ by $(y - k)$ . The shape and the value of $a$ are untouched; only the location changes.

Figure 4: Shifting the vertex from

(0, 0)

(h, k)

moves the whole curve rigidly. The value of

a

, and hence the shape, is unchanged; only the vertex, focus, axis and directrix are translated.

Shifted form	Vertex	Focus	Directrix	Axis
$(y - k)^{2} = 4 a (x - h)$	$(h, k)$	$(h + a, k)$	$x = h - a$	$y = k$
$(y - k)^{2} = - 4 a (x - h)$	$(h, k)$	$(h - a, k)$	$x = h + a$	$y = k$
$(x - h)^{2} = 4 a (y - k)$	$(h, k)$	$(h, k + a)$	$y = k - a$	$x = h$
$(x - h)^{2} = - 4 a (y - k)$	$(h, k)$	$(h, k - a)$	$y = k + a$	$x = h$

Reducing a general equation to standard form

A parabola with axis parallel to a coordinate axis appears as a second-degree equation in which only one variable is squared, for example $4 y^{2} + 12 x - 20 y + 67 = 0$ . Reduce it by completing the square:

Divide through so that the coefficient of the squared variable becomes $1$ .
Group the terms in that variable on one side and complete the square.
Move everything else to the other side and factor out the coefficient of the linear variable, reaching $(y - k)^{2} = 4 a (x - h)$ or its analogue.
Read off the vertex $(h, k)$ , then set $X = x - h, Y = y - k$ and identify $4 a$ , the direction of opening, the focus, the directrix and the latus rectum in the $X Y$ frame.
Translate every answer back by adding $h$ to $X$ and $k$ to $Y$ .

JEE AdvancedWhen the axis is not parallel to a coordinate axis. Given focus

S (x_{0}, y_{0})

and directrix

l x + m y + n = 0

, the definition

S P = P M

gives the equation directly:

(x - x_{0})^{2} + (y - y_{0})^{2} = \frac{( l x + m y + n ) ^{2}}{l ^{2} + m ^{2}} .

Expanding always produces a second-degree equation

A x^{2} + 2 H x y + B y^{2} + 2 G x + 2 F y + C = 0

satisfying

H^{2} = A B

. Conversely, a general second-degree equation represents a parabola precisely when

Δ \neq = 0

and

H^{2} = A B

, where

Δ

is the determinant

Δ = A H G H B F GF C .

Δ = 0

the equation degenerates into a pair of parallel straight lines instead.

4. Parametric Representation

The tidiest way to name a point on $y^{2} = 4 a x$ is with a single parameter $t$ :

x = a t^{2}, y = 2 a t .

These satisfy the equation identically, since $(2 a t)^{2} = 4 a^{2} t^{2} = 4 a \cdot a t^{2}$ . Every real $t$ gives a point on the parabola, and every point on the parabola comes from exactly one $t$ (namely $t = y /2 a$ ). We therefore speak of "the point $t$ ", written $P (t)$ , meaning $P (a t^{2}, 2 a t)$ .

Figure 5: Points of

y^{2} = 4 a x

t = - 2, - 1, 0, 1, 2

. The vertex is

t = 0

, positive

t

gives the upper half and negative

t

the lower half, and

∣ t ∣

grows with distance from the vertex.

Reading the figure: $t = 0$ is the vertex, $t > 0$ sweeps the upper half of the curve and $t < 0$ the lower half, and $∣ t ∣$ grows as the point moves away from the vertex.

Why the parametric form matters: chord, tangent, normal and focal-chord results all collapse into short expressions in

t

, while in

(x, y)

they stay clumsy. Almost every JEE problem on parabolas uses the

t

-parameter at some stage, so it is worth becoming fluent with it now.

5. Focal Distance

The focus-directrix definition gives the focal distance for free, with no distance formula needed. For $P (x_{1}, y_{1})$ on $y^{2} = 4 a x$ , the perpendicular distance to the directrix $x = - a$ is $x_{1} + a$ , so

S P = P M = x_{1} + a .

In parametric form, with $x_{1} = a t^{2}$ ,

S P = a t^{2} + a = a (1 + t^{2}) .

Since $t^{2} \geq 0$ , the smallest focal distance is $a$ , attained at $t = 0$ : the vertex is the point of the parabola nearest to the focus.

Parabola	Focal distance of $(x_{1}, y_{1})$
$y^{2} = 4 a x$	$x_{1} + a$
$y^{2} = - 4 a x$	$a - x_{1}$
$x^{2} = 4 a y$	$y_{1} + a$
$x^{2} = - 4 a y$	$a - y_{1}$

6. Position of a Point Relative to a Parabola

For the parabola $y^{2} = 4 a x$ with $a > 0$ , the position of any point $(x_{1}, y_{1})$ is decided by the sign of

S_{1} = y_{1}^{2} - 4 a x_{1} .

$S_{1} < 0$ : the point lies inside the parabola, on the concave side where the focus is.
$S_{1} = 0$ : the point lies on the parabola.
$S_{1} > 0$ : the point lies outside the parabola.

Figure 6: A point

(x_{1}, y_{1})

lies inside, on, or outside the parabola

y^{2} = 4 a x

according as

S_{1} = y_{1}^{2} - 4 a x_{1}

is negative, zero or positive. The shaded interior is the region containing the focus.

The reasoning is direct. Fix the height $y_{1}$ . The curve meets that horizontal line at $x = y_{1}^{2} /4 a$ , and the interior is everything to the right of it. So the point is inside exactly when $x_{1} > y_{1}^{2} /4 a$ , that is when $y_{1}^{2} - 4 a x_{1} < 0$ .

Adapting the test. Always write the equation as

S = 0

with everything on the left, then substitute the point. For

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

use

S_{1} = (y_{1} - k)^{2} - 4 a (x_{1} - h)

; for

x^{2} = 4 a y

use

S_{1} = x_{1}^{2} - 4 a y_{1}

. "Inside" always means the same side as the focus, whichever way the parabola opens.

7. A Line and a Parabola

Take a line $y = m x + c$ with $m \neq = 0$ and the parabola $y^{2} = 4 a x$ . Substituting for $y$ gives

(m x + c)^{2} = 4 a x ⟹ m^{2} x^{2} + (2 m c - 4 a) x + c^{2} = 0,

a genuine quadratic in $x$ . Its discriminant is

D = (2 m c - 4 a)^{2} - 4 m^{2} c^{2} = 16 a^{2} - 16 am c = 16 a (a - m c) .

Since $a > 0$ , the sign of $D$ is the sign of $(a - m c)$ , and that single quantity settles everything.

Figure 7: Three parallel lines

y = m x + c

with the same slope. The quantity

m c

compared with

a

decides everything:

m c < a

gives a secant,

m c = a

the tangent,

m c > a

no intersection.

Condition	Discriminant	The line meets the parabola in
$m c < a$	$D > 0$	two distinct points (secant)
$m c = a$ , i.e. $c = \frac{a}{m}$	$D = 0$	one point (tangent)
$m c > a$	$D < 0$	no real point

Condition of tangency. The line

y = m x + c

touches

y^{2} = 4 a x

if and only if

c = \frac{a}{m}

(with

m \neq = 0

). Equivalently, the tangent of slope

m

y = m x + \frac{a}{m}

, and it touches the curve at

(\frac{a}{m ^{2}}, \frac{2 a}{m})

Two special lines the condition does not cover.

$m = 0$ (a line parallel to the axis, such as $y = c$ ). Substituting gives $c^{2} = 4 a x$ , a linear equation with the single root $x = c^{2} /4 a$ . So the line meets the parabola in exactly one point, yet it is not a tangent - it cuts straight through the curve. Meeting a conic once is not the same as touching it.
Vertical lines $x = c$ . These have no slope $m$ , so the condition $c = a / m$ cannot be applied. A vertical line meets $y^{2} = 4 a x$ in two points if $c > 0$ , in one point if $c = 0$ (the tangent at the vertex, $x = 0$ ), and in no point if $c < 0$ .

Length of the chord cut by a line

When $m c < a$ the line is a secant. From the quadratic, $∣ x_{1} - x_{2} ∣ = \frac{D}{m ^{2}} = \frac{4 a ( a - m c )}{m ^{2}}$ , and multiplying by $1 + m^{2}$ to convert a horizontal spread into an actual length gives

chord length = \frac{4}{m ^{2}} a (1 + m^{2}) (a - m c) .

8. Chords and Focal Chords

The chord joining two parameters

Let $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ lie on $y^{2} = 4 a x$ . The slope of $P Q$ is

m = \frac{2 a t _{1} - 2 a t _{2}}{a t _{1}^{2} - a t _{2}^{2}} = \frac{2 ( t _{1} - t _{2} )}{( t _{1} - t _{2} ) ( t _{1} + t _{2} )} = \frac{2}{t _{1} + t _{2}},

so the chord, after simplification, is

2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0.

This one equation is the workhorse behind almost every chord result on the parabola. Its length follows from the two coordinate differences $a (t_{1}^{2} - t_{2}^{2})$ and $2 a (t_{1} - t_{2})$ :

∣ P Q ∣ = a ∣ t_{1} - t_{2} ∣ (t_{1} + t_{2})^{2} + 4 .

Focal chord: the condition $t_{1} t_{2} = - 1$

Figure 8: A focal chord

P Q

passes through the focus

S (a, 0)

, and its endpoint parameters always satisfy

t_{1} t_{2} = - 1

. The latus rectum (dashed) is the shortest focal chord, of length

4 a

A chord is a focal chord when it passes through $S (a, 0)$ . Substituting $x = a, y = 0$ into the chord equation gives $2 a + 2 a t_{1} t_{2} = 0$ , so

t_{1} t_{2} = - 1.

Hence if one end of a focal chord has parameter $t$ , the other end has parameter $- \frac{1}{t}$ , and the two extremities are

(a t^{2}, 2 a t) and (\frac{a}{t ^{2}}, - \frac{2 a}{t}) .

Because $t_{1} t_{2} = - 1$ is negative, $t_{1}$ and $t_{2}$ always have opposite signs: the two ends of a focal chord always lie on opposite sides of the axis, which matches the picture.

Length of a focal chord

Adding the two focal distances gives

S P + S Q = a (1 + t^{2}) + a (1 + \frac{1}{t ^{2}}) = a (t + \frac{1}{t})^{2} .

Since $t + \frac{1}{t} \geq 2$ for every real $t \neq = 0$ , the length is at least $4 a$ , with equality when $t = \pm 1$ . That is the latus rectum, confirming that the latus rectum is the shortest focal chord. Written in terms of the angle $α$ that the chord makes with the axis,

ℓ = 4 a csc^{2} α,

which is again least when $α = 9 0^{\circ}$ .

The harmonic mean property

For any focal chord $P S Q$ ,

\frac{1}{S P} + \frac{1}{S Q} = \frac{1}{a ( 1 + t ^{2} )} + \frac{t ^{2}}{a ( 1 + t ^{2} )} = \frac{1}{a} .

Rearranged, $\frac{2 \cdot S P \cdot S Q}{S P + S Q} = 2 a$ : the semi-latus rectum is the harmonic mean of the two segments of any focal chord. This is a favourite JEE result because it holds for every focal chord without exception.

JEE AdvancedCircle on a focal chord as diameter. The circle drawn with any focal chord as diameter touches the directrix. Reason: the distance from the midpoint of

P Q

to the directrix is the average of the distances from

P

and

Q

, which by the definition equals

\frac{1}{2} (S P + S Q) = \frac{1}{2} ∣ P Q ∣

, exactly the radius. By the same argument, the circle on any focal distance

S P

as diameter touches the tangent at the vertex.

Solved Examples

Solved Example 1

Find the vertex, axis, focus, directrix and length of the latus rectum of the parabola

4 y^{2} + 12 x - 20 y + 67 = 0

Solution:

Only $y$ is squared, so the axis is horizontal. Divide by $4$ :

y^{2} + 3 x - 5 y + \frac{67}{4} = 0 ⟹ y^{2} - 5 y = - 3 x - \frac{67}{4} .

Complete the square on the left by adding $(\frac{5}{2})^{2} = \frac{25}{4}$ to both sides:

(y - \frac{5}{2})^{2} = - 3 x - \frac{67}{4} + \frac{25}{4} = - 3 x - \frac{21}{2} = - 3 (x + \frac{7}{2}) .

Put $X = x + \frac{7}{2}, Y = y - \frac{5}{2}$ . Then $Y^{2} = - 3 X$ , which is $Y^{2} = - 4 a X$ with $4 a = 3$ , so $a = \frac{3}{4}$ and the parabola opens to the left.

Vertex: $X = 0, Y = 0 \Rightarrow (- \frac{7}{2}, \frac{5}{2})$ .
Axis: $Y = 0 \Rightarrow y = \frac{5}{2}$ .
Focus: $X = - a, Y = 0 \Rightarrow x = - \frac{3}{4} - \frac{7}{2} = - \frac{17}{4}$ , giving $(- \frac{17}{4}, \frac{5}{2})$ .
Directrix: $X = a \Rightarrow x + \frac{7}{2} = \frac{3}{4} \Rightarrow x = - \frac{11}{4}$ .
Latus rectum: $4 a = 3$ .

Solved Example 2

Find the equation of the parabola whose vertex is

(2, - 3)

and whose focus is

(2, 1)

Solution:

Vertex and focus share the same $x$ -coordinate, so the axis is the vertical line $x = 2$ and the parabola opens upward (the focus is above the vertex).

Here $a =$ distance from vertex to focus $= 1 - (- 3) = 4$ , so $4 a = 16$ and the form is $(x - h)^{2} = 4 a (y - k)$ with $(h, k) = (2, - 3)$ :

(x - 2)^{2} = 16 (y + 3) .

Check: the directrix should be $y = k - a = - 3 - 4 = - 7$ , which is indeed $4$ units below the vertex, mirroring the focus.

Solved Example 3

Find the parametric equations of the parabola

(x - 1)^{2} = - 12 (y - 2)

Solution:

Comparing with $(x - h)^{2} = - 4 a (y - k)$ : $h = 1, k = 2, 4 a = 12 \Rightarrow a = 3$ .

For $X^{2} = - 4 aY$ the parametric point is $X = 2 a t, Y = - a t^{2}$ , that is $X = 6 t, Y = - 3 t^{2}$ . Translating back with $X = x - 1, Y = y - 2$ :

x = 1 + 6 t, y = 2 - 3 t^{2} .

Solved Example 4

Check whether the point

(3, 4)

lies inside or outside the parabola

y^{2} = 4 x

Solution:

Comparing $y^{2} = 4 x$ with $y^{2} = 4 a x$ gives $a = 1$ . Evaluate $S_{1}$ at $(x_{1}, y_{1}) = (3, 4)$ :

S_{1} = y_{1}^{2} - 4 a x_{1} = 16 - 4 \cdot 1 \cdot 3 = 4 > 0.

Since $S_{1} > 0$ , the point $(3, 4)$ lies outside the parabola.

Sanity check: at height $y = 4$ the curve sits at $x = 16/4 = 4$ , and $3 < 4$ , so the point is indeed to the left of the curve, on the outside.

Solved Example 5

Discuss the position of the line

y = x + 1

with respect to the parabola

y^{2} = 4 x

Solution:

Substitute $y = x + 1$ into $y^{2} = 4 x$ :

(x + 1)^{2} = 4 x ⟹ x^{2} - 2 x + 1 = 0 ⟹ (x - 1)^{2} = 0.

The discriminant is zero, so there is exactly one point of intersection: $y = x + 1$ is tangent to $y^{2} = 4 x$ at $(1, 2)$ .

Check with the condition of tangency: here $a = 1, m = 1, c = 1$ , and $\frac{a}{m} = 1 = c$ . The point of contact $(\frac{a}{m ^{2}}, \frac{2 a}{m}) = (1, 2)$ agrees.

Solved Example 6

Prove that the focal distance of the point

P (a t^{2}, 2 a t)

on the parabola

y^{2} = 4 a x

(with

a > 0

) is

a (1 + t^{2})

Solution:

The focus is $S (a, 0)$ and the directrix is $x + a = 0$ . Since $a t^{2} + a > 0$ , the perpendicular distance from $P$ to the directrix is

P M = x_{P} + a = a t^{2} + a = a (1 + t^{2}) .

By the focus-directrix definition, $S P = P M$ , so $S P = a (1 + t^{2})$ .

Compact form: the focal distance of any point $(x_{P}, y_{P})$ on $y^{2} = 4 a x$ is $x_{P} + a$ .

Solved Example 7

t_{1}, t_{2}

are the parameters of the endpoints of a focal chord of

y^{2} = 4 a x

, prove that

t_{1} t_{2} = - 1

Solution:

Let $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ be the endpoints and $S (a, 0)$ the focus. Since $P, S, Q$ are collinear, the slope of $P S$ equals the slope of $P Q$ :

m_{P S} = \frac{2 a t _{1} - 0}{a t _{1}^{2} - a} = \frac{2 t _{1}}{t _{1}^{2} - 1}, m_{P Q} = \frac{2 a t _{1} - 2 a t _{2}}{a t _{1}^{2} - a t _{2}^{2}} = \frac{2}{t _{1} + t _{2}} .

Setting them equal gives $t_{1} (t_{1} + t_{2}) = t_{1}^{2} - 1$ , that is $t_{1}^{2} + t_{1} t_{2} = t_{1}^{2} - 1$ , so

t_{1} t_{2} = - 1.

Note: the step dividing by $(t_{1}^{2} - 1)$ assumes $t_{1} \neq = \pm 1$ . When $t_{1} = \pm 1$ the chord is the latus rectum, and then $t_{2} = \mp 1$ , so $t_{1} t_{2} = - 1$ still holds.

Solved Example 8

Find the length of the focal chord of

y^{2} = 4 a x

whose one endpoint corresponds to the parameter

t

Solution:

By $t_{1} t_{2} = - 1$ the endpoints are $P (a t^{2}, 2 a t)$ and $Q (\frac{a}{t ^{2}}, - \frac{2 a}{t})$ . Their focal distances are

S P = a (t^{2} + 1), S Q = a (\frac{1}{t ^{2}} + 1) = \frac{a ( 1 + t ^{2} )}{t ^{2}} .

Adding,

P Q = S P + S Q = a (1 + t^{2}) (1 + \frac{1}{t ^{2}}) = a (t + \frac{1}{t})^{2} .

By AM-GM, $t + \frac{1}{t} \geq 2$ , so $P Q \geq 4 a$ with equality at $t = \pm 1$ : the latus rectum.

Solved Example 9

If one end of a focal chord of

y^{2} = 16 x

(16, 16)

, find the other end.

Solution:

Here $4 a = 16 \Rightarrow a = 4$ . Find the parameter of $(16, 16)$ from $y = 2 a t$ :

16 = 8 t ⟹ t = 2,

which is consistent with $x = a t^{2} = 4 \cdot 4 = 16$ . The other end has parameter $t^{'} = - \frac{1}{t} = - \frac{1}{2}$ , so

x^{'} = a (t^{'})^{2} = 4 \cdot \frac{1}{4} = 1, y^{'} = 2 a t^{'} = 8 \cdot (- \frac{1}{2}) = - 4.

The other end is $(1, - 4)$ .

Check: the focus is $(4, 0)$ ; the slope from $(16, 16)$ to $(4, 0)$ is $\frac{16}{12} = \frac{4}{3}$ , and from $(4, 0)$ to $(1, - 4)$ it is $\frac{- 4}{- 3} = \frac{4}{3}$ . Collinear, as required.

Solved Example 10

P S Q

is a focal chord of

y^{2} = 4 a x

(with

a > 0

) and

S

is the focus, prove that

\frac{1}{P S} + \frac{1}{S Q} = \frac{1}{a}

Solution:

Take the endpoints as $P (a t^{2}, 2 a t)$ and $Q (\frac{a}{t ^{2}}, - \frac{2 a}{t})$ . Then

P S = a (1 + t^{2}), S Q = a (1 + \frac{1}{t ^{2}}) = \frac{a ( t ^{2} + 1 )}{t ^{2}} .

Adding the reciprocals,

\frac{1}{P S} + \frac{1}{S Q} = \frac{1}{a ( 1 + t ^{2} )} + \frac{t ^{2}}{a ( 1 + t ^{2} )} = \frac{1 + t ^{2}}{a ( 1 + t ^{2} )} = \frac{1}{a} .

Equivalently $\frac{2 \cdot P S \cdot S Q}{P S + S Q} = 2 a$ , so the semi-latus rectum $2 a$ is the harmonic mean of the two segments of any focal chord.

Solved Example 11

Find the set of values of

λ

for which the line

y = 3 x + λ

meets the parabola

y^{2} = 4 x

at two distinct points.

Solution:

Here $a = 1, m = 3, c = λ$ . The line is a secant when $m c < a$ :

3 λ < 1 ⟹ λ < \frac{1}{3} .

So $λ \in (- \infty, \frac{1}{3})$ .

Caution: the step from $m c < a$ to $c < a / m$ is only valid because $m = 3$ is positive. If $m$ were negative, dividing by $m$ would reverse the inequality, so always work with $m c < a$ and divide only at the end.

Solved Example 12

Find the midpoint of the chord that the line

x + y = 2

cuts on the parabola

y^{2} = 4 x

Solution:

Substitute $y = 2 - x$ into $y^{2} = 4 x$ :

(2 - x)^{2} = 4 x ⟹ x^{2} - 8 x + 4 = 0.

The discriminant is $64 - 16 = 48 > 0$ , so the line really is a secant. By the sum of roots, $x_{1} + x_{2} = 8$ , so the midpoint has $x$ -coordinate $\frac{x _{1} + x _{2}}{2} = 4$ .

The midpoint lies on the line as well, so $y = 2 - 4 = - 2$ . The midpoint is $(4, - 2)$ .

Check: $S_{1} = (- 2)^{2} - 4 (4) = - 12 < 0$ , so the midpoint is inside the parabola, exactly as a chord midpoint must be.

Solved Example 13

JEE AdvancedFind the equation of the parabola whose focus is

(2, 3)

and whose directrix is

x - 4 y + 3 = 0

Solution:

Apply the definition $S P = P M$ directly. For a point $P (x, y)$ ,

(x - 2)^{2} + (y - 3)^{2} = \frac{∣ x - 4 y + 3∣}{1 ^{2} + ( - 4 ) ^{2}} = \frac{∣ x - 4 y + 3∣}{17} .

Squaring and clearing the denominator,

17 [(x - 2)^{2} + (y - 3)^{2}] = (x - 4 y + 3)^{2} .

Expanding the left side gives $17 x^{2} + 17 y^{2} - 68 x - 102 y + 221$ , and the right side gives $x^{2} + 16 y^{2} + 9 - 8 x y + 6 x - 24 y$ . Collecting everything on the left:

16 x^{2} + 8 x y + y^{2} - 74 x - 78 y + 212 = 0.

Verification: comparing with $A x^{2} + 2 H x y + B y^{2} + \dots$ gives $A = 16, B = 1, H = 4$ , so $H^{2} = 16 = A B$ . The condition for a parabola is satisfied, and the presence of the $x y$ term tells us the axis is not parallel to either coordinate axis.

Common Mistakes to Avoid

Watch out

Mixing up $a$ and $4 a$ . The standard form is $y^{2} = 4 a x$ , not $y^{2} = a x$ . Given $y^{2} = 8 x$ , we get $4 a = 8 \Rightarrow a = 2$ , so the focus is $(2, 0)$ , not $(8, 0)$ , and the latus rectum is $8$ , not $2$ .
Assuming "inside" means "above" or "to the left". Inside always means on the same side as the focus. For $y^{2} = - 4 a x$ the focus is $(- a, 0)$ , so the inside is the region to the left of the curve.
Applying $S_{1} = y_{1}^{2} - 4 a x_{1}$ to a shifted parabola. That form of the test is only for $y^{2} = 4 a x$ . For $(y - k)^{2} = 4 a (x - h)$ you must use $S_{1} = (y_{1} - k)^{2} - 4 a (x_{1} - h)$ .
Treating "meets in one point" as "is a tangent". A line parallel to the axis ( $m = 0$ , e.g. $y = 3$ ) cuts $y^{2} = 4 a x$ in exactly one point but is not a tangent. The tangency condition $c = a / m$ silently assumes $m \neq = 0$ .
Dividing $m c < a$ by a negative $m$ without flipping the sign. The safe statement of the secant condition is $m c < a$ . Convert it to $c < a / m$ only after checking that $m > 0$ .
Reading off the focus with the wrong sign. For $y^{2} = 4 a x$ the focus is $(+ a, 0)$ ; for $y^{2} = - 4 a x$ it is $(- a, 0)$ . When an equation appears as $y^{2} = 4 A x$ with $A$ possibly negative, first decide the direction of opening, then place the focus $∣ A ∣$ units from the vertex in that direction.
Using $t_{1} t_{2} = - 1$ for a general chord. This holds only for focal chords. A general chord joining $t_{1}$ and $t_{2}$ satisfies no such restriction.
Forgetting to check that the chord is real. Before computing a midpoint or a chord length, confirm $D > 0$ . A "midpoint" computed from a line that never meets the curve is meaningless.

Frequently Asked Questions

Q1. What is the standard equation of a parabola?

The standard equation of a right-opening parabola with vertex at the origin is $y^{2} = 4 a x$ , where $a$ is the distance from the vertex to the focus. Its focus is $(a, 0)$ , directrix is $x + a = 0$ , axis is the $x$ -axis, and latus rectum has length $4 a$ . Three other standard orientations exist: $y^{2} = - 4 a x$ (opens left), $x^{2} = 4 a y$ (opens up), $x^{2} = - 4 a y$ (opens down).

Q2. What is the latus rectum of a parabola?

The latus rectum is the chord of the parabola that passes through the focus and is perpendicular to the axis. For $y^{2} = 4 a x$ its equation is $x = a$ , its length is $4 a$ , and its endpoints are $L (a, 2 a)$ and $L^{'} (a, - 2 a)$ . It is the shortest possible focal chord.

Q3. What are the parametric equations of a parabola?

For $y^{2} = 4 a x$ , every point can be written as $(a t^{2}, 2 a t)$ for some real $t$ . The parameter $t$ acts as a single coordinate: $t = 0$ is the vertex, $t > 0$ gives the upper half and $t < 0$ the lower half. Chord, tangent, normal and focal-chord results all take their simplest form in terms of parameters.

Q4. How do you find the focal distance of a point on a parabola?

For a point $(x_{1}, y_{1})$ on $y^{2} = 4 a x$ , the focal distance equals $x_{1} + a$ . In parametric form, if the point is $P (a t^{2}, 2 a t)$ , the focal distance is $a (1 + t^{2})$ . This follows directly from the focus-directrix definition, since the perpendicular distance from any point on the curve to the directrix $x = - a$ equals $x_{1} + a$ . For $x^{2} = 4 a y$ the corresponding result is $y_{1} + a$ .

Q5. What is the condition for a line to touch a parabola?

The line $y = m x + c$ with $m \neq = 0$ is tangent to $y^{2} = 4 a x$ if and only if $c = \frac{a}{m}$ . Equivalently, the tangent of slope $m$ is $y = m x + \frac{a}{m}$ , touching the parabola at $(\frac{a}{m ^{2}}, \frac{2 a}{m})$ . A line with $m = 0$ meets the curve exactly once but is a secant, not a tangent.

Q6. What is a focal chord and what is special about it?

A focal chord is any chord of the parabola that passes through the focus. Its most useful property: if the endpoints have parameters $t_{1}$ and $t_{2}$ , then $t_{1} t_{2} = - 1$ , which also means the two ends lie on opposite sides of the axis. Two further results: the length of a focal chord making angle $α$ with the axis is $4 a csc^{2} α$ , and the semi-latus rectum $2 a$ is the harmonic mean of its two segments.

Q7. How do you determine whether a point lies inside or outside a parabola?

For the parabola $y^{2} = 4 a x$ , evaluate $S_{1} = y_{1}^{2} - 4 a x_{1}$ at the given point $(x_{1}, y_{1})$ . If $S_{1} < 0$ the point is inside (the same side as the focus), if $S_{1} = 0$ it lies on the parabola, and if $S_{1} > 0$ it is outside. For a shifted parabola $(y - k)^{2} = 4 a (x - h)$ , use $S_{1} = (y_{1} - k)^{2} - 4 a (x_{1} - h)$ .

Q8. Why is the sum of reciprocals of focal chord segments equal to $1/ a$ ?

For any focal chord $P S Q$ of $y^{2} = 4 a x$ , we have $P S = a (1 + t^{2})$ and $S Q = a (1 + 1/ t^{2})$ , where $t$ and $- 1/ t$ are the parameters of $P$ and $Q$ . Adding the reciprocals gives $\frac{1}{P S} + \frac{1}{S Q} = \frac{1}{a}$ . This is equivalent to saying the semi-latus rectum $2 a$ is the harmonic mean of $P S$ and $S Q$ , a defining metric property of the parabola.

Q9. How is a parabola different from an ellipse or a hyperbola?

The parabola is the borderline conic between ellipse and hyperbola. It has eccentricity exactly $1$ , one focus, one directrix, one unbounded branch, no centre and no asymptote. An ellipse (with $e < 1$ ) is closed and has two foci; a hyperbola (with $e > 1$ ) has two disconnected branches, two foci and two asymptotes. All parabolas are similar to one another, differing only in scale through the value of $a$ .

Q10. How can you tell whether a general second-degree equation represents a parabola?

Write the equation as $A x^{2} + 2 H x y + B y^{2} + 2 G x + 2 F y + C = 0$ . It represents a parabola when $H^{2} = A B$ and the determinant $Δ$ of the matrix with rows $(A, H, G)$ , $(H, B, F)$ , $(G, F, C)$ is non-zero. If $H = 0$ as well, only one variable is squared and the axis is parallel to a coordinate axis, so the curve can be reduced to standard form by completing the square. If $Δ = 0$ the equation degenerates into a pair of parallel lines rather than a parabola.

Previous year questions on Introduction to Parabola

9 questions from past papers, each with a step-by-step solution.

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1. Definition and the Standard Equation

Deriving y2=4ax from the definition

2. Elements of the Standard Parabola

The four standard forms

Vocabulary you will need

3. Shifted Parabola: Vertex Not at the Origin

Reducing a general equation to standard form

4. Parametric Representation

5. Focal Distance

6. Position of a Point Relative to a Parabola

7. A Line and a Parabola

Length of the chord cut by a line

8. Chords and Focal Chords

The chord joining two parameters

Focal chord: the condition t1​t2​=−1

Length of a focal chord

The harmonic mean property

Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the standard equation of a parabola?

Q2. What is the latus rectum of a parabola?

Q3. What are the parametric equations of a parabola?

Q4. How do you find the focal distance of a point on a parabola?

Q5. What is the condition for a line to touch a parabola?

Q6. What is a focal chord and what is special about it?

Q7. How do you determine whether a point lies inside or outside a parabola?

Q8. Why is the sum of reciprocals of focal chord segments equal to 1/a?

Q9. How is a parabola different from an ellipse or a hyperbola?

Q10. How can you tell whether a general second-degree equation represents a parabola?

Previous year questions on Introduction to Parabola

Ready to master Parabola?

Deriving $y^{2} = 4 a x$ from the definition

Focal chord: the condition $t_{1} t_{2} = - 1$

Q8. Why is the sum of reciprocals of focal chord segments equal to $1/ a$ ?