Chapter 10 · Conic Sections

10.2 The Parabola

A parabola is the set of points equally far from a fixed point (the focus) and a fixed line (the directrix). With the focus at $(a,0)$ and the directrix $x=-a$ it is $y^2=4ax$, and its latus rectum has length $4a$.

Look at a satellite dish on a rooftop. Signals from a satellite $36\,000\ \text{km}$ away arrive as parallel rays; the dish has to send every one of them to the small receiver held out in front of it on a short arm. A spherical bowl would not do it — rays hitting near the rim would miss the receiver. The shape that works perfectly is a parabola spun about its axis, and the receiver sits at a special point called its focus. Run the rays backwards and you have a car headlamp or a torch: a bulb at the focus sends out a parallel beam. Parabolas also describe the path of a cricket ball or a water jet (if we ignore air resistance), the cables of a suspension bridge and many concrete arches.

In Class 10 you met parabolas as graphs of quadratic polynomials $y=ax^2+bx+c$. Here we start instead from a geometric rule about distances — every point of a parabola is as far from the focus as from a fixed line, the directrix — and turn it into the standard equation $y^2=4ax$. From that equation you will read off the focus, the directrix, the axis and the latus rectum, and you will build the equation of a parabola from given information.

Focus, Directrix and the Standard Equation

📘 Definition 10.4 — Parabola

A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) of the plane. The fixed line is the directrix and the fixed point $S$ is the focus. The line through the focus perpendicular to the directrix is the axis of the parabola, and the point where the parabola meets its axis is the vertex of the parabola. The chord through the focus perpendicular to the axis is the latus rectum.

The vertex lies on the axis halfway between the focus and the directrix: it is the one point of the axis equally far from both. Why must the focus not lie on the directrix? If it did, the points equidistant from both would be just the line through the focus perpendicular to the directrix — a straight line, not a curve.

To get the simplest possible equation, choose the axes cleverly. Let the distance from the focus to the directrix be $2a$ ($a\gt0$). Put the vertex at the origin, the axis of the parabola along the $x$-axis and the focus on the positive side: then the focus is $S(a,0)$ and the directrix is the line $x=-a$.

📐 Theorem 10.3 — Standard equation of a parabola

The parabola with focus $S(a,0)$ ($a\gt0$) and directrix $x=-a$ has equation $$y^2=4ax.$$

Proof. Let $P(x,y)$ be any point and let $M$ be the foot of the perpendicular from $P$ to the directrix, so $M=(-a,y)$ and $PM=|x+a|$. By definition $P$ is on the parabola exactly when $PS=PM$, i.e. $$\sqrt{(x-a)^2+y^2}=|x+a|.$$ Squaring (both sides are non-negative): $x^2-2ax+a^2+y^2=x^2+2ax+a^2$, which simplifies to $y^2=4ax$.
Conversely, suppose $y^2=4ax$. Then $x=\tfrac{y^2}{4a}\ge0$, and $PS^2=(x-a)^2+4ax=(x+a)^2$; since $x+a\gt0$, $PS=x+a=PM$. So every point satisfying the equation is on the parabola, and the equation describes the parabola exactly. $\blacksquare$

The proof gives a useful by-product: for a point $P(x,y)$ on $y^2=4ax$, its distance from the focus is $PS=x+a$. Placing the focus to the left, above or below the vertex instead gives three more standard forms. In each, the vertex is the origin, the latus rectum has length $4a$, and the non-squared variable names the axis of the parabola.

EquationFocusDirectrixAxisOpensEnds of latus rectum
$y^2=4ax$$(a,0)$$x=-a$$y=0$right$(a,\pm2a)$
$y^2=-4ax$$(-a,0)$$x=a$$y=0$left$(-a,\pm2a)$
$x^2=4ay$$(0,a)$$y=-a$$x=0$up$(\pm2a,a)$
$x^2=-4ay$$(0,-a)$$y=a$$x=0$down$(\pm2a,-a)$

To read a parabola in standard form: see which variable is squared ($y^2$ means the axis is the $x$-axis), and look at the sign of the other side (positive means the parabola opens towards the positive direction of that axis). Then $4a$ is the size of the coefficient, the focus is $a$ from the vertex on the open side, and the directrix is $a$ from the vertex on the other side.

✏️ Example 1 — Reading off a parabola
Find the focus, the axis, the directrix and the length of the latus rectum of
(a) $y^2=12x$;   (b) $x^2=-16y$;   (c) $3y^2=-8x$.
(a) $y^2=4ax$ with $4a=12$, $a=3$: focus $(3,0)$, axis the $x$-axis, directrix $x=-3$, latus rectum $12$. It opens to the right.
(b) $x^2=-4ay$ with $4a=16$, $a=4$: focus $(0,-4)$, axis the $y$-axis, directrix $y=4$, latus rectum $16$. It opens downwards.
(c) Divide by $3$: $y^2=-\tfrac83x$, so $4a=\tfrac83$ and $a=\tfrac23$: focus $\left(-\tfrac23,0\right)$, axis the $x$-axis, directrix $x=\tfrac23$, latus rectum $\tfrac83$. It opens to the left.
Check (a) The end $(3,6)$ of the latus rectum: $6^2=36=12\cdot3$ ✓, and it is $6$ from the focus and $3+3=6$ from the directrix ✓.
Interactive 1 — Focus, Directrix and the Reflector

The point $P$ runs along $y^2=4ax$. The red segment $PS$ goes to the focus and the blue segment $PM$ goes straight across to the directrix: watch the readouts — they are always equal. Switch on the rays to see the reflecting property of a dish: every ray coming in parallel to the axis bounces off the curve and passes through the focus. Change $a$ to make the parabola wider or narrower.

1.5
$P(x,y)$—
$PS$—
$PM=x+a$—
$y^2$ and $4ax$—

The Latus Rectum and Equations from Given Data

📐 Theorem 10.4 — Latus rectum of a parabola

The latus rectum of $y^2=4ax$ ($a\gt0$) has ends $(a,2a)$ and $(a,-2a)$, so its length is $4a$. The same length $4a$ holds for each of the other three standard forms.

Proof. The latus rectum lies on the line $x=a$ through the focus, perpendicular to the axis. Putting $x=a$ in $y^2=4ax$ gives $y^2=4a^2$, so $y=\pm2a$. The ends are $(a,\pm2a)$ and the distance between them is $2a-(-2a)=4a$. The other forms are reflections of this one in a coordinate axis or in the line $y=x$, which do not change lengths. $\blacksquare$

The latus rectum gives a quick, accurate sketch: plot the vertex and the two ends of the latus rectum, which lie level with the focus at distance $2a$ on either side of the axis, and draw a smooth curve through the three points, opening away from the directrix. It also tells you how "wide" the parabola is — a large $a$ means a wide, flat parabola, and a small $a$ a narrow one.

🧭 Finding the equation of a parabola (vertex at the origin)

1. Decide the axis and the direction of opening: the focus lies on the open side; the directrix on the other side. A point on the curve also tells you the side.
2. Choose the matching standard form: $y^2=\pm4ax$ (axis along the $x$-axis) or $x^2=\pm4ay$ (axis along the $y$-axis).
3. Find $a$ from the focus or directrix (its distance from the vertex), from the latus rectum ($4a$), or by substituting a known point.
4. Check one point, or the focus–directrix distance $2a$.

✏️ Example 2 — From the focus and directrix
Find the equation of the parabola with
(a) focus $(6,0)$ and directrix $x=-6$;   (b) focus $(0,-3)$ and directrix $y=3$.
(a) The focus is on the positive $x$-axis and the directrix is $x=-6$: the form is $y^2=4ax$ with $a=6$, so $y^2=24x$.
(b) The focus is on the negative $y$-axis, so the parabola opens downwards: $x^2=-4ay$ with $a=3$, so $x^2=-12y$.
Check (b) The point $(6,-3)$ satisfies $36=-12(-3)$ ✓; its distance from the focus is $6$ and from the line $y=3$ is $6$ ✓.
✏️ Example 3 — From the vertex and one more fact
Find the equation of the parabola with vertex $(0,0)$ and
(a) focus $(-2,0)$;   (b) axis along the $x$-axis, passing through $(-3,6)$;   (c) axis along the $y$-axis and latus rectum of length $6$, opening upwards.
(a) The focus is $2$ units to the left of the vertex: $y^2=-4ax$ with $a=2$, i.e. $y^2=-8x$.
(b) Axis along the $x$-axis: $y^2=kx$ for some $k$. Substituting $(-3,6)$: $36=-3k$, $k=-12$. So $y^2=-12x$ (it opens to the left, as a point with negative $x$ requires).
(c) $4a=6$ and the form $x^2=4ay$: $x^2=6y$.
Check (b) $a=3$, focus $(-3,0)$: the point $(-3,6)$ is an end of the latus rectum, $2a=6$ above the focus ✓.
✏️ Example 4 — Straight from the definition
Find the equation of the parabola with focus $S(2,3)$ and directrix $y=-1$, and find its vertex.
Idea The vertex is not at the origin, so no standard form applies directly. Go back to Definition 10.4: $PS=PM$.
Distances For $P(x,y)$: $PS^2=(x-2)^2+(y-3)^2$, and the distance from $P$ to the line $y=-1$ is $|y+1|$.
Simplify $(x-2)^2+(y-3)^2=(y+1)^2$ gives $(x-2)^2=(y+1)^2-(y-3)^2=8y-8$, i.e. $$(x-2)^2=8(y-1).$$
Vertex The vertex is midway between $S(2,3)$ and the directrix along the axis $x=2$: the point $(2,1)$, which is where the right side is $0$. The parabola is $x^2=8y$ (with $a=2$) moved so that its vertex is at $(2,1)$ — and indeed $S$ is $a=2$ above the vertex ✓.

Parabolas at Work

A parabolic dish or mirror has a reflecting property: every ray arriving parallel to the axis of the parabola is reflected through the focus, and every ray leaving the focus is reflected parallel to the axis. (The reason uses the angle a ray makes with the curve, which needs tangents to curves — you will meet them in Class 12 and in JEE. Here we use the property.) This is why the receiver of a dish antenna, the bulb of a headlamp and the cooking pot of a solar cooker are all placed at the focus. To design one, put the vertex at the origin and the axis along the $y$-axis, so the cross-section is $x^2=4ay$; a point on the rim then gives $a$.

✏️ Example 5 — Where does the receiver go?
A dish antenna is $2.4\ \text{m}$ across at its rim and $0.3\ \text{m}$ deep at its centre. Its cross-section is a parabola. How far from the vertex should the receiver be placed?
Model Vertex at the origin, axis up: $x^2=4ay$. The rim is $1.2\ \text{m}$ either side of the axis and $0.3\ \text{m}$ above the vertex, so $(1.2,\,0.3)$ is on the curve.
Solve $1.2^2=4a(0.3)$: $1.44=1.2a$, so $a=1.2$.
Answer The focus is $(0,1.2)$: the receiver goes $1.2\ \text{m}$ from the vertex along the axis.
Check The rim point is $\sqrt{1.2^2+(0.3-1.2)^2}=\sqrt{1.44+0.81}=1.5\ \text{m}$ from the focus and $0.3+1.2=1.5\ \text{m}$ from the directrix $y=-1.2$ ✓.
✏️ Example 6 — A parabolic arch
A concrete arch has the shape of a parabola with a vertical axis. It is $40\ \text{m}$ wide at its base and $10\ \text{m}$ high at the centre. Find its height $5\ \text{m}$ from one end of the base.
Model Put the vertex (top of the arch) at the origin; the arch opens downwards: $x^2=-4ay$. The base ends are $(\pm20,-10)$.
Find $a$ $400=-4a(-10)=40a$, so $a=10$ and the arch is $x^2=-40y$.
Height $5\ \text{m}$ from an end means $x=\pm15$: $y=-\tfrac{225}{40}=-5.625$. That point is $5.625\ \text{m}$ below the top, so its height above the base is $10-5.625=4.375\ \text{m}$.
Check Heights fall off as the square of the distance from the centre: at $15\ \text{m}$ from the centre the drop is $\left(\tfrac{15}{20}\right)^2\times10=5.625\ \text{m}$ ✓.
Interactive 2 — The Four Standard Forms

Pick a standard form and a value of $a$. The panel draws the parabola with its focus $S$, directrix, axis and latus rectum, and the readouts list what you would write in an answer. Press Play to cycle through the four forms: notice that the axis always goes with the variable that is not squared, and the sign decides the side it opens towards.

1
equation—
focus—
directrix—
latus rectum—
⚠️ Common mistakes with parabolas

1. Taking $4a$ for $a$. In $y^2=12x$ the focus is $(3,0)$, not $(12,0)$: first write the coefficient as $4a$, then divide by $4$.

2. The wrong axis. In $x^2=8y$ it is $x$ that is squared, yet the axis is the $y$-axis. The axis goes with the variable that is not squared.

3. Focus and directrix on the same side. The focus is inside the curve and the directrix outside it, on opposite sides of the vertex: for $y^2=-8x$ the focus is $(-2,0)$ and the directrix is $x=+2$.

4. Latus rectum $2a$. Its ends are $2a$ from the axis on each side, so its full length is $4a$.

5. Using a standard form when the vertex is not at the origin. For a focus such as $(2,3)$ with directrix $y=-1$, use $PS=PM$ directly (Example 4).

🚀 Beyond the board syllabus (JEE Advanced)

In JEE Advanced 2026 (parametric equations and foci of conics); not in JEE Main 2026 or the CBSE year-end exam. JEE Advanced uses two short-cuts for $y^2=4ax$ all the time.

Parametric form. For every real $t$, the point $(at^2,\,2at)$ lies on $y^2=4ax$, since $(2at)^2=4a\cdot at^2$; and every point of the parabola arises from exactly one $t$, namely $t=\tfrac{y}{2a}$. So one letter $t$ describes a point — "the point $t$".

Focal distance. From the proof of Theorem 10.3, the distance of $P(x,y)$ on the parabola from the focus is $$PS=x+a=a(1+t^2).$$ For example, the ends of the latus rectum ($t=\pm1$) are $2a$ from the focus.

Worked example. Find the points of $y^2=12x$ whose focal distance is $7$. Solution. $a=3$, so $x+3=7$, $x=4$, and $y^2=48$, $y=\pm4\sqrt3$: the points $(4,\pm4\sqrt3)$, i.e. $t=\pm\tfrac{2}{\sqrt3}$. Check: $\sqrt{(4-3)^2+48}=\sqrt{49}=7$ ✓.

Exercises

I
Find the focus, the axis, the directrix and the length of the latus rectum.
  • 1 $y^2=20x$
  • 2 $x^2=8y$
  • 3 $y^2=-6x$
  • 4 $x^2=-20y$
  • 5 $5y^2=16x$
  • 6 $2x^2+9y=0$
  • 7 Find the ends of the latus rectum of $y^2=16x$.
  • 8 Which way does $x^2=-y$ open, and where is its focus?
1 $a=5$: focus $(5,0)$, axis $y=0$, directrix $x=-5$, latus rectum $20$.
2 $a=2$: focus $(0,2)$, axis $x=0$, directrix $y=-2$, latus rectum $8$.
3 $4a=6$, $a=\tfrac32$: focus $\left(-\tfrac32,0\right)$, axis $y=0$, directrix $x=\tfrac32$, latus rectum $6$.
4 $a=5$: focus $(0,-5)$, axis $x=0$, directrix $y=5$, latus rectum $20$.
5 $y^2=\tfrac{16}{5}x$, $a=\tfrac45$: focus $\left(\tfrac45,0\right)$, axis $y=0$, directrix $x=-\tfrac45$, latus rectum $\tfrac{16}{5}$.
6 $x^2=-\tfrac92y$, $4a=\tfrac92$, $a=\tfrac98$: focus $\left(0,-\tfrac98\right)$, axis $x=0$, directrix $y=\tfrac98$, latus rectum $\tfrac92$.
7 $a=4$: ends $(4,8)$ and $(4,-8)$.
8 $x^2=-4ay$ with $4a=1$: it opens downwards and its focus is $\left(0,-\tfrac14\right)$.
II
Find the equation of the parabola (vertex at the origin in every part).
  • 1 Focus $(4,0)$, directrix $x=-4$.
  • 2 Focus $(0,-5)$, directrix $y=5$.
  • 3 Focus $\left(0,\tfrac72\right)$.
  • 4 Focus $(-5,0)$.
  • 5 Axis along the $x$-axis, passing through $(2,-4)$.
  • 6 Axis along the $y$-axis, passing through $(-4,-2)$.
  • 7 Directrix $y=\tfrac32$.
  • 8 Axis along the $y$-axis, opening upwards, latus rectum of length $10$.
  • 9 Focus $(3,0)$; give also the ends of its latus rectum.
Check: is your point on the parabola, and is $PS=PM$?
left vs right side—
$PS$—
$PM$—
1 $y^2=16x$.
2 $x^2=-20y$.
3 $a=\tfrac72$, opening upwards: $x^2=14y$.
4 $y^2=-20x$.
5 $y^2=kx$: $16=2k$, $k=8$: $y^2=8x$.
6 $x^2=ky$: $16=-2k$, $k=-8$: $x^2=-8y$.
7 The directrix is above the vertex, so the parabola opens downwards with $a=\tfrac32$: $x^2=-6y$.
8 $4a=10$: $x^2=10y$.
9 $a=3$: $y^2=12x$; ends of the latus rectum $(3,\pm6)$.
III
Using the definition $PS=PM$.
  • 1 Find the parabola with focus $(0,0)$ and directrix $x=-4$. Where is its vertex?
  • 2 Find the parabola with focus $(2,3)$ and directrix $y=1$, and its vertex.
  • 3 Find the parabola with focus $(1,-1)$ and directrix $x=3$, and its vertex.
  • 4 Show that $(9,6)$ lies on $y^2=4x$, and verify that it is equally far from the focus and the directrix.
  • 5 For which $a$ does $y^2=4ax$ pass through $(3,-6)$? Give its focus.
  • 6 Find the points of $y^2=8x$ that are $10$ units from the focus.
  • 7 A chord of $y^2=4ax$ through the vertex makes an angle of $45^\circ$ with the axis. Find its length.
1 $x^2+y^2=(x+4)^2$ gives $y^2=8x+16=8(x+2)$. Vertex $(-2,0)$, midway between the focus and the directrix.
2 $(x-2)^2+(y-3)^2=(y-1)^2$ gives $(x-2)^2=(y-1)^2-(y-3)^2=4y-8=4(y-2)$. Vertex $(2,2)$.
3 $(x-1)^2+(y+1)^2=(x-3)^2$ gives $(y+1)^2=(x-3)^2-(x-1)^2=-4x+8=-4(x-2)$. Vertex $(2,-1)$; it opens to the left.
4 $6^2=36=4\cdot9$ ✓. Focus $(1,0)$: $PS=\sqrt{64+36}=10$; directrix $x=-1$: $PM=9+1=10$ ✓.
5 $36=12a$, $a=3$; focus $(3,0)$.
6 $a=2$ and $PS=PM=x+2=10$, so $x=8$ and $y^2=64$: the points $(8,8)$ and $(8,-8)$.
7 The chord lies on $y=x$ (or $y=-x$). Then $x^2=4ax$ gives $x=0$ or $x=4a$: the ends are $(0,0)$ and $(4a,4a)$, so the length is $4a\sqrt2$.
IV
Applications (vertex at the origin, axis vertical, unless told otherwise).
  • 1 A car headlamp reflector is $20\ \text{cm}$ across and $6\ \text{cm}$ deep, with a parabolic cross-section. How far from the vertex should the bulb be?
  • 2 A parabolic arch is $30\ \text{m}$ wide at the base and $9\ \text{m}$ high. Find its equation (vertex at the top) and its height $6\ \text{m}$ from the centre.
  • 3 The cable of a suspension bridge hangs in a parabola between two towers $100\ \text{m}$ apart. It is $20\ \text{m}$ above the road at the towers and $5\ \text{m}$ above it at the middle. Find the height of the cable above the road $20\ \text{m}$ from the middle.
  • 4 A fountain jet rises to a greatest height of $2\ \text{m}$ at a horizontal distance of $1.5\ \text{m}$ from its nozzle (at ground level). Taking the top of the jet as the vertex, find the equation of the jet and its height $0.5\ \text{m}$ horizontally from the nozzle.
  • 5 The receiver of a dish $2.4\ \text{m}$ across is $0.9\ \text{m}$ from the vertex. How deep is the dish?
  • 6 A mirror has cross-section $y^2=24x$ (units cm), with its axis horizontal. Where do rays arriving parallel to the axis meet after reflection?
1 $x^2=4ay$ through $(10,6)$: $100=24a$, $a=\tfrac{25}{6}\approx4.17\ \text{cm}$ from the vertex.
2 $x^2=-4ay$ through $(15,-9)$: $225=36a$, $a=\tfrac{25}{4}$, so $x^2=-25y$. At $x=6$: $y=-\tfrac{36}{25}=-1.44$; height $9-1.44=7.56\ \text{m}$.
3 Vertex at the lowest point of the cable: $x^2=4ay$, and the tower tops are $15\ \text{m}$ above it at $x=\pm50$: $2500=60a$, $a=\tfrac{125}{3}$. At $x=20$: $y=\tfrac{400}{4a}=\tfrac{400\cdot3}{500}=2.4$; height above the road $5+2.4=7.4\ \text{m}$.
4 $x^2=-4ay$ through the nozzle $(-1.5,-2)$: $2.25=8a$, $a=\tfrac{9}{32}$, so $x^2=-\tfrac98y$. $0.5\ \text{m}$ from the nozzle is $x=-1$: $y=-\tfrac89$; height $2-\tfrac89=\tfrac{10}{9}\approx1.11\ \text{m}$.
5 $x^2=3.6y$ through $(1.2,d)$: $1.44=3.6d$, $d=0.4\ \text{m}$.
6 $4a=24$, $a=6$: at the focus $(6,0)$, $6\ \text{cm}$ from the vertex along the axis.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask you to read off the focus, axis, directrix and latus rectum of a standard parabola (1–2 marks) and to find its equation from the focus, the directrix or a point (2–3 marks); case studies use arches, dishes and headlamps. JEE adds the focal distance $x+a$ and the parametric point $(at^2,2at)$.

E1. Board · 1 mark The focus of the parabola $x^2=-12y$ is (A) $(0,3)$   (B) $(0,-3)$   (C) $(-3,0)$   (D) $(0,-12)$

Idea $x^2=-4ay$ with $4a=12$, $a=3$: it opens downwards and the focus is $(0,-a)=(0,-3)$.
Trap (C) puts the focus on the wrong axis; (D) uses $4a$ instead of $a$; (A) ignores the minus sign.
Answer (B).

E2. Board · 2 marks Find the equation of the parabola with vertex at the origin and axis along the $x$-axis that passes through $(-2,4)$. Write down its focus.

Form $y^2=kx$; substituting $(-2,4)$: $16=-2k$, $k=-8$. (1 mark)
Answer $y^2=-8x$, with $4a=8$, $a=2$: focus $(-2,0)$. (The given point is an end of the latus rectum.) (1 mark)

E3. Board · 3 marks A parabola has focus $(0,3)$ and directrix $y=-3$. Find its equation, the length and the ends of its latus rectum, and the points where it meets the line $y=12$.

Equation Vertex at the origin, opening upwards, $a=3$: $x^2=12y$. (From $PS=PM$: $x^2+(y-3)^2=(y+3)^2$ gives the same.) (1 mark)
Latus rectum Length $4a=12$; ends $(\pm6,3)$. (1 mark)
Answer $y=12$: $x^2=144$, $x=\pm12$, so the points are $(12,12)$ and $(-12,12)$. (1 mark)

E4. Board · Case study The entrance of a railway tunnel is a parabolic arch $24\ \text{m}$ wide at road level and $8\ \text{m}$ high at the centre, with a vertical axis. Take the top of the arch as the origin. (i) Find the equation of the arch. (1 mark) (ii) Find the height of the arch $4\ \text{m}$ from the centre line. (1 mark) (iii) A container truck is $6\ \text{m}$ tall. What is the greatest width it can have if it is to pass under the arch? (2 marks)

(i) $x^2=-4ay$ through $(12,-8)$: $144=32a$, $a=\tfrac92$, so $x^2=-18y$. (1 mark)
(ii) $x=4$: $y=-\tfrac{16}{18}=-\tfrac89$; height $8-\tfrac89=\tfrac{64}{9}\approx7.11\ \text{m}$. (1 mark)
(iii) set-up A height of $6\ \text{m}$ is $2\ \text{m}$ below the top: $y=-2$, so $x^2=36$, $x=\pm6$. (1 mark)
Answer At $6\ \text{m}$ the arch is $12\ \text{m}$ wide, so the truck (centred) must be less than $12\ \text{m}$ wide. (1 mark)

E5. JEE Main The parabola $y^2=4ax$ passes through $(2,-6)$. The distance between its focus and its directrix is (A) $\tfrac92$   (B) $9$   (C) $18$   (D) $36$

Idea $36=8a$, so $a=\tfrac92$. The focus is $a$ on one side of the vertex and the directrix $a$ on the other: distance $2a=9$.
Trap (A) is $a$ itself (vertex to focus); (C) is the latus rectum $4a$.
Answer (B).

E6. JEE Advanced Numerical answer (to two decimal places). $P$ is a point in the first quadrant on $y^2=12x$ whose distance from the focus $S$ is $12$. If $O$ is the vertex, find the area of triangle $OSP$.

Focal distance $a=3$ and $PS=x+3=12$ (🚀 box), so $x=9$, $y^2=108$, $y=6\sqrt3$ (first quadrant).
Area Base $OS=3$ along the axis, height $6\sqrt3$: area $=\tfrac12\cdot3\cdot6\sqrt3=9\sqrt3$.
Answer $9\sqrt3\approx15.59$.
⏱ Fast route In parametric form $P=(3t^2,6t)$ and $PS=3(1+t^2)=12$ gives $t=\sqrt3$, so the height is $6\sqrt3$ at once.
🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. Is the path of a water jet really a parabola — and where is its focus?

Why it matters. Engineers who design fountains, fire-fighting nozzles and irrigation sprinklers need to know where the water lands. If air resistance is small, every particle of water moves like a thrown ball and its path is a parabola with a vertical axis. Testing this with real data is exactly how scientists check a mathematical model.

What to do.

  1. Point a garden hose (or squeeze a water bottle with a small hole in the cap) at a gentle upward angle in front of a wall with a grid of tape marks every $10\ \text{cm}$, or hold a ruler in the picture for scale. Photograph the jet side-on with your phone, keeping the camera level.
  2. On the photo, mark the highest point of the jet as the origin. Read the coordinates $(x,y)$ of $6$–$8$ points along the jet, with $y$ measured downwards as negative.
  3. If the jet is a parabola $x^2=-4ay$, then $\dfrac{x^2}{-y}$ should be the same number $4a$ for every point. Compute it for each point in a spreadsheet.
  4. Take the mean as $4a$, and mark the focus $(0,-a)$ and the directrix $y=a$ on the photo. Check one point: is it equally far from both?

Tools: A phone camera, a hose or bottle, tape or a ruler for scale, a spreadsheet or a GeoGebra-style app

Hand in: The photo with your points, focus and directrix marked; your table of $x$, $y$ and $x^2/(-y)$; your value of $a$; and a paragraph on where the model fits well and where it fails (look near the end of the jet, where the stream breaks into drops).

Going further. Change the pressure (open the tap more) without changing the angle. How does $a$ change? Physics says $4a=\dfrac{2u^2\cos^2\theta}{g}$ for launch speed $u$ and angle $\theta$; use your value of $a$ to estimate the launch speed.

Instructor note: typical garden-hose jets give $x^2/(-y)$ constant to within about $10\%$ over the first metre. The ratio usually drifts near the end of the jet because of air drag and break-up into droplets — a good discussion of model limits. Photos taken at an angle distort distances; insist on a side-on, level camera.

🧪 Self-check quiz

0/5 correct

Q1. The directrix of $y^2=8x$ is

Q2. The length of the latus rectum of $x^2=-10y$ is

Q3. The parabola with vertex $(0,0)$ and focus $(0,4)$ is

Q4. Which parabola opens to the left?

Q5. The point $(4,4)$ lies on $y^2=4x$. Its distance from the focus is

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