# EQUATION OF PARABOLA GIVEN FOCUS AND DIRECTRIX

## About "Equation of parabola given focus and directrix"

Equation of parabola given focus and directrix :

Here we are going to see how to find equation of the parabola if focus and directrix is given.

Example 1 :

Find the equation of the parabola whose focus and directrix is (-1, -2) and x - 2y + 3 = 0

Solution :

Let P (x, y) be any point on the parabola. If PM is drawn perpendicular to the directrix respectively.

FP/PM = e

Since it is parabola e = 1

FP = PM

√(x +1)² + (y+ 2)² = (x-2y+3)/(√1²+2²)

√(x +1)² + (y+ 2)² = (x-2y+3)/(√5)

taking squares on both sides

(x+1)² + (y+2)² = (x-2y+3)²/5

5[x²+2x+1+y²+4y+4] = x²+4y²+9-4xy-12y+6x

5x²- x² + 4xy + 5y²- 4y²+ 10x - 6x + 20y + 12y + 25-9 = 0

4x²+ 4xy + y²+ 4x + 32y + 16 = 0

Example 2 :

Find the equation of the parabola whose focus and directrix is (2, -3) and y - 2 = 0 respectively

Solution :

Let P (x, y) be any point on the parabola. If PM is drawn perpendicular to the directrix.

FP/PM = e

Since it is parabola e = 1

FP = PM

√(x-2)² + (y+3)² = (y-2)/(√0²+1²)

√(x-2)² + (y+3)² = (y-2)/1

taking squares on both sides

(x-2)² + (y+3)² = (y-2)²

x² - 4 x + 4 + y² + 6 y + 9 = y² - 4y + 4

x² + y² - y²- 4 x + 6 y + 4 y + 9 + 4 - 4 = 0

x² - 4 x + 10 y + 9  = 0

Example 3 :

Find the equation of the parabola whose focus and directrix is (2, -3) and 2y - 3 = 0 respectively

Solution :

Let P (x, y) be any point on the parabola. If PM is drawn perpendicular to the directrix.

FP/PM = e

Since it is parabola e = 1

FP = PM

√(x-2)² + (y+3)² = (2y-3)/(√0²+2²)

√(x-2)² + (y+3)² = (2y-3)/√4

taking squares on both sides

(x-2)² + (y+3)² = (2y-3)²/4

4[x² - 4 x + 4 + y² + 6 y + 9] = 4y² - 12y + 9

4x² + 4y² - 4y² - 16x + 24y + 12y + 52 - 9 = 0

4x² - 16x + 36y + 43 = 0

Example 4 :

Find the equation of the parabola whose focus and directrix is (-1, 3) and 2x + 3y = 3 respectively

Solution :

Let P (x, y) be any point on the parabola. If PM is drawn perpendicular to the directrix.

FP/PM = e

Since it is parabola e = 1

FP = PM

√(x+1)² + (y-3)² = (2x+3y-3)/(√2²+3²)

√(x+1)² + (y-3)² = (2x+3y-3)/√13

taking squares on both sides

(x+1)² + (y-3)² = (2x+3y-3)²/13

13[x²+2x+1+y²-6y+9] =  4x²+9y²+9+12xy-18y-12x

13x²-4x²-12xy+13y²-9y²+26x+12x-78y+18y+130-9 = 0

9 x²-12 x y + 4 y²+ 38 x - 60 y + 121 = 0

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