# Examples on Parabola

**Example 1**

For the following parabolas, find coordinates of the focus and the equation of the directrix

(i) y

^{2 }= 8x (ii) x

^{2 }= 6y (iii) y

^{2 }= -12x (iv) x

^{2 }= 16y

**Solution**

- We compare y
^{2 }= 8x with y^{2 }= 4ax, so that 4a = 8 or a = 2.

Coordinates of the focus = (a, 0) = (2, 0).

Equation of the directrix: x+a = 0 or x+2 = 0.

- We compare x
^{2 }= 6y with x^{2 }= 4ay so that 4a = 6 or a = 3/2.

Coordinates of the focus = (0, a)=(0,3/2).

Equation of the directrix : y + a = 0 or y + 3/2 = 0 or 2y + 3 = 0.

- We compare y
^{2 }= -12x with y^{2 }= -4ax, so that 4a = 12 or a = 3.

Coordinates of the focus = (-a, 0) = (-3, 0).

Equation of the directrix: x = a or x = 3

- We compare x
^{2 }= 16y with x^{2}= 4ay, so that 4a = 16 or a = 4.

Coordinates of the focus = (0, a) = (0, 4)

Equation of the directrix: y = a or y = 4.

**Example 2**

Find the equation of the parabola with vertex at the origin and satisfying the condition

- focus at (- a, 0)
- directrix y = 2
- passing through (2, 3) and axis along x-axis
- directrix x + 3 = 0.

**Solution**

- Let M be the foot of the perpendicular from F(-a,0) to the directrix

Let the coordinates of M be (x , y ).

Then

â‡’ x' = a and y' = 0.

The coordinates of M are (a, 0).

Equation of the axis is y = 0 and equation of the directrix is x = a.

Let P(x, y) be any point on the parabola. Let T be the foot of perpendicular from P to the directrix. By definition, PF = PT

But

Since PF = PT, we have PF

^{2}= PT

^{2}

â‡’ (x + a)

^{2}+ y

^{2}= (x - a)

^{2}â‡’ y

^{2 }= (x - a)

^{2 }- (x + a)

^{2 }= -4ax.

So equation of the required parabola is y

^{2 }= -4ax.

- Let M be the foot of perpendicular from the origin to the directrix y = 2.

Let the coordinates of the focus be F(x', y'). Since the vertex is mid-point of MF,

â‡’ x' = 0, y' = -2

Thus, coordinates of F are (0, -2)

Let P(x, y) be any point on the parabola, then

PF = PM' [see in figure given below.]

â‡’ PF

^{2 }= PM'

^{2}

â‡’ x

^{2 }+ (y + 2)

^{2 }=

â‡’ x

^{2 }= (y - 2)

^{2 }- (y + 2)

^{2 }= -8y

Thus, equation of the required parabola is x

^{2 }= -8y.

- Equation of the parabola with vertex at the origin and axis along x-axis is y
^{2 }= 4ax.

â‡’ a = 8a or .

Thus, equation of the required parabola is or 2y

^{2 }= 9x.

- Let M be the foot of perpendicular from the vertex (origin) to the directrix x+3 = 0. Then coordinates of M are (âˆ’ 3,0). Let F (x', y') be the focus of the parabola. Then since origin is the mid-point of MF.

So the coordinates of F are (3, 0).

If P(x, y) is any point on the parabola, then

PF = PM' â‡’ PF

^{2 }= PM'

^{2}

â‡’ y

^{2 }= (x + 3)

^{2 }- (x - 3)

^{2 }â‡’ y

^{2 }= 12x.

Thus, equation of the required parabola is y

^{2 }= 12x.