Table of Contents

Last modified on October 15th, 2024

A parabola is a U-shaped curve in which all points are equidistant from a fixed point and a fixed straight line. The point is the **focus** of the parabola, and the line is the **directrix**.

The focus lies on the axis of symmetry, and the directrix is parallel to either the x-axis or the y-axis. However, the focus never lies on the directrix.

The focus and the directrix for the standard equations of a parabola centered at (0, 0) are:

- Focus is at
**(a, 0)** - The equation of the directrix is
**x + a = 0**

- Focus is at
**(-a, 0)** - The equation of the directrix is
**x – a = 0**

- Focus is at
**(0, a)** - The equation of the directrix is
**y + a = 0**

- Focus is at
**(0, -a)** - The equation of the directrix is
**y – a = 0**

- Focus is at
**(h, k + a)** - Directrix is
**y = k – a**

**Note**: If the parabola is in vertex form y = a(x – h)^{2} + k, then the focus is at ${\left( h,k+\dfrac{1}{4a}\right)}$ and directrix is ${y=k-\dfrac{1}{4a}}$

- Focus is at
**(h + a, k)** - Directrix is
**x = h – a**

**Note**: If the parabola is in the vertex form as x = a(y – k)^{2} + h, then the focus is at ${\left( h+\dfrac{1}{4a},k\right)}$ and directrix is ${x=h-\dfrac{1}{4a}}$

Let us find the focus and the directrix of the parabola (x – 3)^{2} = -8(y – 5)

Given, (x – 3)^{2} = -8(y – 5) …..(i)

The given equation is of the form: (x – h)^{2} = 4a(y – k) …..(ii)

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

a = -2, h = 3, and k = 5

Here, the focus is at (h, k + a) = (3, 5 – 2) = (3, 3)

The directrix is y = k – a

⇒ y = 5 – (-2)

⇒ y = 7

Thus, the focus is at (3, 3), and the directrix is y = 7

**Find the focus and directrix of the parabola y ^{2} = 8x**

Solution:

Given, y^{2} = 8x …..(i)

The given equation is of the form: y^{2} = 4ax …..(ii)

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

a = 2

Here, the focus is at (a, 0) = (2, 0)

The equation of the directrix is x + a = 0 ⇒ x + 2 = 0 ⇒ x = -2

Thus, the focus is at (2, 0), and the directrix is x = -2

**Find the vertex, focus, and directrix of the parabola given by the equation (x – 2) ^{2} = 12(y – 3)**

Solution:

Given, (x – 2)^{2} = 12(y – 3) …..(i)

The given equation is of the form: (x – h)^{2} = 4a(y – k) …..(ii)

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

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

Here, the vertex is at (h, k) = (2, 3)

The focus is at (h, k + a) = (2, 3 + 3) = (2, 6)

The directrix is y = k – a

⇒ y = 3 – 3

⇒ y = 0

Thus, the vertex is at (2, 3), the focus is at (2, 6), and the directrix is y = 0

**Determine the focus and directrix of y = -(x – 2) ^{2} – 4**

Solution:

Given, y = -(x – 2)^{2} – 4 …..(i)

⇒ (y + 4) = -(x – 2)^{2}

⇒ (x – 2)^{2} = – (y + 4) …..(ii)

The given equation is of the form: (x – h)^{2} = 4a(y – k) …..(iii)

Comparing the equations (ii) and (iii), we have

a = ${-\dfrac{1}{4}}$, h = 2, and k = -4

Here, the focus is at (h, k + a) = ${\left( 2,-4-\dfrac{1}{4}\right)}$ = (2, -4.25)

The directrix is y = k – a

⇒ y = ${-4-\left( -\dfrac{1}{4}\right)}$

⇒ y = -3.75

Thus, the focus is at (2, -4.25), and the directrix is y = -3.75

**Problem – **Finding the equation of the parabola when **FOCUS** and **DIRECTRIX **are known

Let F be the focus, and the line MN represents the directrix.

Now, a perpendicular line FG is drawn from F on the directrix and bisect it at V.

Then, VF = VG ⇒ Distance of V from the focus = Distance of V from the directrix

By the definition of the parabola, since V lies on the parabola, FG = 2a ⇒ VF = VG = a

Let us consider V as the vertex. Then, VG is perpendicular to the directrix and parallel to the x-axis.

The coordinates of the focus F are (h, k), and the equation of the directrix MN is x = b.

Since PB is perpendicular to the directrix x = b, and point B is with the coordinates (b, y)

Let us consider a point P(x, y) on the parabola. Now, FP and PB are joined.

As we know, P lies on the parabola.

Now, FP = PB

⇒ FP^{2} = PB^{2}

⇒ (x – h)^{2} + (y – k)^{2} = (x – b)^{2} + (y – y)^{2}

⇒ x^{2} – 2hx + h^{2} + (y – k)^{2} = x^{2} – 2bx + b^{2}

On adding (2hx – b^{2}) on both sides, we get

⇒ x^{2} – 2hx + h^{2} + 2hx – b^{2} + (y – k)^{2} = x^{2} – 2bx + b^{2} + 2hx – b2

⇒ 2(h – b)x = (y – k)^{2} + h^{2} – b^{2}

On dividing equation by 2(h – b), we get

${x=\dfrac{\left( y-k\right) ^{2}}{2\left( h-b\right) }+\dfrac{h^{2}-b^{2}}{2\left( h-b\right) }}$

⇒ ${x=\dfrac{\left( y-k\right) ^{2}}{2\left( h-b\right) }+\dfrac{\left( h+b\right) }{2}}$ …..(i)

Similarly, when directrix y = b, we get

${y=\dfrac{\left( x-h\right) ^{2}}{2\left( k-b\right) }+\dfrac{\left( k+b\right) }{2}}$ …..(ii)

When V is the origin, VF is the x-axis of length a. Then, the coordinates of F will be (a, 0), and the directrix MN is x = -a.

h = a, k = 0 and b = -a

Using the equation (i), we get

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

⇒ ${x=\dfrac{y^{2}}{4a}}$

⇒ y^{2} = 4ax

**Derive the equation of a parabola with focus (0, 2) and directrix y = −2**

Solution:

The vertex is the midpoint between the focus and directrix, which is (0,0). The distance from the vertex to the focus is a = 2Using the formula y^{2} = 4ax, the equation of the parabola is y^{2} = 8x

Last modified on October 15th, 2024