Table of Contents

Last modified on September 28th, 2024

Let us imagine a basketball game. As a player shoots the ball, the ball follows a curved path before landing in the hoop. The path that the ball takes is a parabola.

A parabola is a U-shaped curve that is also found in the path of water in a fountain or the shape of satellite dishes.

Mathematically, it is the graph of a quadratic function in two variables, y = ax^{2 }+ bx + c, that can be defined as the set of all points that are equidistant from a fixed point, called the **focus**, and a fixed line, called the **directrix**. The vertex is the point where the parabola is closest to the directrix and where it changes direction. The y = ax^{2 }+ bx + c is thus a parabolic function.

here,

P is any point on parabola

F is the focus

PB = PF

The shape of a parabola is found in the trajectory of rockets and the arc of water from the fountains. It has many applications, such as in headlights, bridge cables, and solar panels.

**Vertex:**It is the turning point of the parabola and also the highest or lowest point, depending on whether it opens upward or downward. The vertex of the above parabola is at (0, 0).**Axis of Symmetry:**It is the dashed line passing through the vertex of the parabola and dividing it into two symmetric halves.**Focus:**Has 1 focus, with coordinates (0, a)**Directrix:**It is the line drawn perpendicular to the axis of the parabola. Here, the line parallel to the x-axis and passing through the point (0, -a) is the directrix.**Focal Chord:**It is the chord that intersects the parabola at two distinct points and passes through the focus.**Latus Rectum:**It is the focal chord of the parabola that is perpendicular to its axis and passes through the focus. The endpoints of the latus rectum are (2a, a) and (-2a, a), and its length is 4a.**Focal Distance:**It is the distance from a point (x_{1}, y_{1}) on the parabola to the focus. The focal distance is also equal to the perpendicular distance from this point to the directrix.**Eccentricity:**It is the ratio of the distance between any point on the curve and the focus to the distance between that point and the directrix. For a parabola, the eccentricity is always equal to 1.

The equation of a parabola depends on its orientation, the position of its vertex, and whether it is centered at the origin or elsewhere.

A **horizontal parabola** is a parabola that opens sideways, either to the left or to the right.

Here,

- The equation of the axis of symmetry is y = 0
- The equation of directrix is x = -a
- Vertex at (0, 0)
- Focus at (a, 0)
- The length of the latus rectum is 4a

This parabola opens to the right, forming a ‘⊂’ shaped curve.

**Derivation: y ^{2} = 4ax**

Let us derive the above equation.

Let us consider a point P on the parabola with the coordinates (x, y)

By the definition of a parabola, the distance of point P from the focus F and the distance of point P from the directrix is the same.

Now, we will draw a perpendicular line from point P that intersects the directrix at B, as shown.

As we know, the eccentricity of a parabola is always 1

We have e = 1

⇒ ${\dfrac{PF}{PB}=1}$

⇒ PF = PB

Since the coordinates of the focus are F(a,0), we can determine its distance from the point P (x, y)

Using the coordinate distance formula, we get

PF = ${\sqrt{\left( x-a\right) ^{2}+\left( y-0\right) ^{2}}}$ = ${\sqrt{\left( x-a\right) ^{2}+y^{2}}}$

Also, the equation of the directrix is x + a = 0. Thus, by using the perpendicular distance formula, we get

PB = ${\dfrac{x+a}{\sqrt{1^{2}+0^{2}}}}$ = ${x+a}$

Now, PF = PB

${\sqrt{\left( x-a\right) ^{2}+y^{2}}=x+a}$

By squaring the equation on both sides,

(x – a)^{2} + y^{2} = (x + a)^{2}

⇒ x^{2} – 2ax + a^{2} + y^{2} = x^{2} + 2ax + a^{2}

⇒ y^{2} – 2ax = 2ax

⇒ y^{2} = 4ax

If the parabola is of the form y^{2} = 4ax, then its parametric coordinates are (at^{2}, 2at). It represents all the points on the parabola.

Here,

- The equation of the axis of symmetry is y = 0
- The equation of directrix is x = a
- Vertex at (0, 0)
- Focus at (-a, 0)
- The length of the latus rectum is 4a

This parabola opens to the left, forming a ‘⊃’ shaped curve.

When a parabola** **opens either upward or downward, it is called a **vertical parabola**.

Here,

- The equation of the axis of symmetry is x = 0
- The equation of directrix is y = -a
- Vertex at (0, 0)
- Focus at (0, a)
- The length of the latus rectum is 4a

This parabola opens upward, forming a ‘U’ shaped curve.

Here,

- The equation of the axis of symmetry is x = 0
- The equation of directrix is y = a
- Vertex at (0, 0)
- Focus at (0, -a)
- The length of the latus rectum is 4a

This parabola opens downward, forming a ‘⋂’ shaped curve.

As we derived the equation for the parabola y^{2} = 4ax, we can similarly derive the equations for other types of parabolas.

Below is a summary of the formulas to determine the axis, directrix, vertex, focus, and length of the latus rectum for parabolas in different orientations when expressed in standard form:

**Find the focus from the given equation of the parabola: ****a) y ^{2} = 20x**

Solution:

a) Given y^{2} = 20x …..(i)

As we know, the standard equation of the parabola is y^{2} = 4ax …..(ii)

Comparing the equations (i) and (ii),

a = 5

Thus, the focus is at (a, 0) = (5, 0)

b) Given x^{2} = -4y …..(i)

As we know, the standard equation of the parabola is x^{2} = -4ay …..(ii)

Comparing the equations (i) and (ii),

a = 1

Thus, the focus is at (0, -a) = (0, -1)

**The equation of a parabola is y ^{2 }= 16x. Find the length of the latus rectum, focus, and vertex.**

Solution:

As we know,

The equation of a parabola is y^{2 }= 16x

Now, 4a = 16

a = 4

Thus,

The length of the latus rectum = 4a = 4 × 4 = 16

Focus = (a, 0) = (4, 0)

Vertex = (0, 0)

Another useful form of writing the equation of a parabola is the vertex form, which highlights the vertex of the parabola when graphed.

The vertex form of the equation of a **vertical parabola** is given by:

Here,

- The equation of the axis of symmetry is x = h
- The equation of directrix is ${y=k-\dfrac{1}{4a}}$
- Vertex at (h, k)
- Focus at ${\left( h,k+\dfrac{1}{4a}\right)}$
- The length of the latus rectum is ${\dfrac{1}{a}}$

**Derivation of Quadratic Function**: **y = ax ^{2} + bx + c**

By rearranging the vertex form of the parabola, we get

y = a(x – h)^{2} + k

⇒ y = a(x^{2} – 2hx + h^{2}) + k

⇒ y = ax^{2} – 2ahx + ah^{2} + k

⇒ y = ax^{2} + (-2ah)x + (ah^{2} + k)

If we replace (-2ah) with b and (ah^{2} + k) with c, the standard form of a parabola is also written as:

**y = ax**^{2}** + bx + c**, here, a ≠ 0, b, and c are constants.

The value of ‘a’ determines whether the parabola opens upward or downward:

- If a > 0, then the parabola opens upward.
- If a < 0, then the parabola opens downward.

The vertex form of the equation of a **horizontal** **parabola** is given by:

Here,

- The equation of the axis of symmetry is y = k
- The equation of directrix is ${x=h-\dfrac{1}{4a}}$
- Vertex at (h, k)
- Focus at ${\left( h+\dfrac{1}{4a},k\right)}$
- The length of the latus rectum is ${\dfrac{1}{a}}$

Here is a table listing all the formulas that are used to find the axis, directrix, vertex, focus, and length of the latus rectum of the parabolas whose vertex is at (h, k):

As we obtained the quadratic form for the parabola y = ax^{2} + bx + c, we can similarly derive the equations for vertical parabolas.

The quadratic equation of a vertical parabola is written as **x = ay**^{2}** + by + c**. Here, a ≠ 0, b, and c are constants.

To graph a parabola, we find the vertex of the parabola and the axis of symmetry, and then, sketch the curve.

For the equation of the parabola y = ax^{2} + bx + c, the x-coordinate for the vertex is ${h=-\dfrac{b}{2a}}$

By substituting this value in the equation, the y-coordinate for the vertex is: k = a(h)^{2} + b(h) + c

Let us graph the parabola y = x^{2}

Comparing the given equation with the standard form of a parabola y = ax^{2} + bx + c, we get

a = 1, b = 0, and c = 0

Since a is positive, the parabola opens up.

The vertex can be calculated as:

${h=-\dfrac{b}{2a}}$ = ${-\dfrac{0}{2\times 1}}$ = ${0}$

Now, substituting h in the original equation y = x^{2}, we get

k = f(h) = 0^{2} = 0

The vertex is (0, 0)

The length of the latus rectum = ${\dfrac{1}{a}}$ = ${1}$

Focus = ${\left( h,k+\dfrac{1}{4a}\right)}$ = ${\left( 0,0+\dfrac{1}{4\times 1}\right)}$ = ${\left( 0,0.25\right)}$

The axis of symmetry is x = 0

Directrix is ${y=k-\dfrac{1}{4a}}$

⇒ ${y=0-\dfrac{1}{4\times 1}}$

⇒ ${y=-0.25}$

Now, plotting the given quadratic equation y = x^{2} on the graph, we get

**Graph the parabola x = 3(y – 2) ^{2} + 4. Find the focus, vertex, and length of the latus rectum.**

Solution:

Given x = 3(y – 2)^{2} + 4 …..(i)

As we know, the standard form of the parabola is x = a(y – k)^{2} +h …..(ii)

Comparing the equations (i) and (ii), we get

a = 3, h = 4, and k = 2

Since a is positive, the parabola opens to the right.

Now,

The vertex is at (h, k) = (4, 2)

The length of the latus rectum = ${\dfrac{1}{a}}$ = ${\dfrac{1}{3}}$

Focus = ${\left( h+\dfrac{1}{4a},k\right)}$ = ${\left( 4+\dfrac{1}{4\times 3},2\right)}$ = ${\left( \dfrac{49}{12},2\right)}$

The axis of symmetry is y = k ⇒ y = 2

Directrix is ${x=h-\dfrac{1}{4a}}$

⇒ ${x=4-\dfrac{1}{4\times 3}}$

⇒ ${x=\dfrac{47}{12}}$

Now, plotting the given quadratic equation y = x^{2} on the graph, we get the required graph.

**Find the equation of the parabola whose graph is shown below.**

Solution:

The graph has a vertex at (2, 3). Hence, the equation of the parabola in vertex form is written as

y = a(x – 2)^{2} + 3

We now use the y-intercept at (0,−1) to find the coefficient a

-1 = a(0 – 2) + 3

Now, we will solve for ‘a’

a = 2

Thus, the equation of the parabola, whose graph is shown above, is

y = 2(x – 2)^{2} + 3

A line that touches the parabola at a single point is called its tangent.

If the equation of the parabola is y^{2} = 4ax, the equation of its tangent at the point of contact (x_{1}, y_{1}) is yy_{1} = 2a(x + x_{1})

The line drawn perpendicular to the tangent that passes through both the point of contact and the focus of the parabola is called the normal.

If the parabola is of the form y^{2} = 4ax and the normal passes through the point (x_{1}, y_{1}) with a slope of ${m=-\dfrac{y_{1}}{2a}}$, then its equation is ${\left( y-y_{1}\right) =-\dfrac{y_{1}}{2a}\left( x-x_{1}\right)}$

The chord of contact is formed by joining the points of contact of the tangents drawn from an external point to the parabola.

If a point (x_{1}, y_{1}) is outside the parabola, the equation of the chord of contact is yy_{1} = 2x(x + x_{1})

Here,

- P is any point with coordinates (x
_{1}, y_{1}) - CD and EF are tangents
- AB is chord of contact
- BG is normal

**Problem: **Finding the equation of a parabola when the **VERTEX** and **DIRECTRIX** are given

**If the vertex of the parabola is at (3, 2) and the directrix of the parabola is y = 5, find the equation of the parabola.**

Solution:

The equation of the parabola in the vertex form is generally written as (x – h)^{2} = 4p(y – k) …..(i)

Here,

(h, k) = vertex of the parabola

p = distance from the vertex to the directrix (or the focus)

Given the vertex is (3, 2) and the directrix is y = 5

Here,

h = 3, k = 2, and p = 2 – 5 = -3

Substituting the values in the equation (i), we get

(x – 3)^{2} = 4(-3)(y – 2)

⇒ x^{2} – 6x + 9 = -12y + 24

⇒ 12y = -x^{2} + 6x – 9 + 24

⇒ 12y = -x^{2} + 6x + 15

⇒ ${y=-\dfrac{1}{12}x^{2}+\dfrac{6}{12}x+\dfrac{15}{12}}$

⇒ ${y=-\dfrac{1}{12}x^{2}+\dfrac{1}{2}x+\dfrac{5}{4}}$

Thus, the equation of the parabola is ${y=-\dfrac{1}{12}x^{2}+\dfrac{1}{2}x+\dfrac{5}{4}}$

Last modified on September 28th, 2024