Loading....
Coupon Accepted Successfully!

 

Examples on Parabola


Example 1
For the following parabolas, find coordinates of the focus and the equation of the directrix
(i) y2 = 8x               (ii) x2 = 6y            (iii) y2 = -12x             (iv) x2 = 16y

Solution
  1. We compare y2 = 8x with y2 = 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.
  1. We compare x2 = 6y with x2 = 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.
  1. We compare y2 = -12x with y2 = -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
  1. We compare x2 = 16y with x2 = 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
  1. focus at (- a, 0)
  2. directrix y = 2
  3. passing through (2, 3) and axis along x-axis
  4. directrix x + 3 = 0.
Solution
  1. Let M be the foot of the perpendicular from F(-a,0) to the directrix  
Then V is the mid-point of FM.
Let the coordinates of M be (x , y ).
Then
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 PF2 = PT2
(x + a)2 + y2 = (x - a)2 y2 = (x - a)2 - (x + a)2 = -4ax.
So equation of the required parabola is y2 = -4ax.
 
  1. Let M be the foot of perpendicular from the origin to the directrix y = 2.
Coordinates of M are (0, 2).
Let the coordinates of the focus be F(x', y'). Since the vertex is mid-point of MF,

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.]


PF2 = PM'2
x2 + (y + 2)2 =
x2 = (y - 2)2 - (y + 2)2 = -8y
Thus, equation of the required parabola is x2 = -8y.
  1. Equation of the parabola with vertex at the origin and axis along x-axis is y2 = 4ax.
Since, it passes through (2, 3), we must have 9 = 4a(2)
a = 8a or .
Thus, equation of the required parabola is or 2y2 = 9x.

  1. Let M be the foot of perpendicular from the vertex (origin) to the directrix x+3 = 0. Then coordinates of M are ( 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' PF2 = PM'2

y2 = (x + 3)2 - (x - 3)2 y2 = 12x.
Thus, equation of the required parabola is y2 = 12x.




Test Your Skills Now!
Take a Quiz now
Reviewer Name