Solved Examples

Example 1:

Find the equation of the parabola whose focus is (3, –4) and directrix is the line parallel to 6x – 7y + 9 = 0 and directrix passes through point (3/2,2).

Solution:

   Let (x, y) be any point on the parabola. Then by definition, the distance between (x, y) and the focus (3, – 4) must be equal to the length of perpendicular from (x, y) on directrix. So first we will find the equation of the directrix.

   The line parallel to 6x – 7y + 9 = 0 is

   6x – 7y + 6 = 0 …… (1)

   Since directrix passes through (3/2,2), this point wil satisfy equation 91) hence

   6 (3/2) – 7 (2) + k = 0

⇒ K = – 9 + 14 = 5.

   Equation of directrix is 6x – 7y + 5 = 0

Now by definition of parabola,

   √({(x-3)²+(y+4)² } )= (6x-7y+5)/√(6²+7² )

⇒ 85 {(x – 3)² + (y + 4)²} = (6x – 7y + 5)²

   ⇒ 49x² + 36y² + 84xy – 570x + 750y + 2100 = 0

Example 2:

   Find the equation of the parabola whose directrix makes an isosceles right angled triangle of area 4 square units with the axis in the 3rd quadrant and focus is on the line y = x, 2 units away from the origin.

Solution:

First we find the equation of directrix. Let the directrix form the isosceles triangle OAB with OA = OB = a.

   parabola-directix-sol

Then according to the given condition,

(Δ OAB) = 4

   ⇒ 1/2 a² = 4

⇒ a = ± 2√2 {∵the triangle in 3^rd quadrant∴=-2√2}

Therefore the co-ordinate, of A and B are (–2√2, 0) and (0, –2√2) respectively.

So, equation of directrix

(y – 0) = ((0+2√2)/(-2√2-0)) (x + 2√2) ⇒ x + y + 2√2 = 0

   Now the focus S is on line y = x and 2 units away from the origin i.e.

OS = 2 ⇒ point (√2, √2) by definition of parabola, we have

√((x-√2)²+(y-√2)² )=|(x+y+2√2)/√(1²+1² )|

   ⇒ x² + y² – 8√2x – 8√2y – 2xy = 0

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