In this page Equation of parabola, we are going to see how to derive equation of the required parabola from the given focus, vertex and directrix.

**Finding the equation of the parabola whose focus and directrix is given:**

**Example 1:**

Find the equation of the parabola whose focus is the point (2,1) and whose directrix is the straight line 2x+y+1=0.

**Solution:**

Let P(x,y) be a point on the parabola. S is the given focus (2,1).

The given directrix is 2x+y+1=0. If PM is drawn perpendicular to the directrix.

Since we have parabola, the eccentricity is 1.

So SP/PM=1

SP²= PM²

(x-2)²+(y-1)² = [(2x+y+1)/√(2²+1²)}]²

x²-4x+4+y²-2y+1 = [(2x+y+1)²]/5

Cross multiplying,

5x²+5y²-20x-10y+25 = 4x²+y²+1+4xy+2y+4x

Simplifying

** x²-4xy+4y²-24x-12y+24=0**

**Example 2:**

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

**Solution:**

Let P(x,y) be a point on the parabola. S is the given focus (3,-2).

The given directrix is x-y+3=0. If PM is drawn perpendicular to the directrix.

Since we have parabola, the eccentricity is 1.

So SP/PM=1

SP²= PM²

(x-3)²+(y+2)² = [(x-y+3)/√(1²+1²)}]²

x²-6x+9+y²+2y+4 = [(x-y+1)²]/2

Cross multiplying,

2x²+2y²-12x+4y+26 = x²+y²+1-2xy-2y+2x

Simplifying

** x²+2xy+y²-14x+6y+25=0**

** Finding the equation of parabola whose vertex and foci are given:**

**Example 3:**

Find the equation of the parabola whose focus is (2,6) and vertex is (2,2)

**Solution:**

In the given focus and vertex, the x coordinates are same. So the focus and vertex are one above the another.And the parabola is a vertical parabola whose x coordinate is squared.

Since the focus is above the vertex, the value of 'a' is positive and that is the distance between the y coordinates.

a = 6-2 = 4.

The equation of the parabola in the vertex form is

(x-h)² = 4a(y-k)

(x-2)² = 4(4)(y-2)

**(x-2)² = 16(y-2)**

If we simplify further

x²+4-2x=16y-32

** x²-2x-16y+36=0**

**Finding the equation of the parabola whose vertex and directrix is given:**

**Example 4:**

Find the equation of the parabola whose vertex is (3,-1) and the equation of the directrix is y= -3.

**Solution:**

The given directrix is y=-3 which is horizontal line. Since the directrix is perpendicular to the axis of symmetry, then the given parabola is a vertical parabola whose x coordinate is squared.

The distance between the vertex and directrix is ∣-3-(-1)∣ = 2.

Since the directrix is below the vertex, the parabola opens up. So the value of 'a' is positive. which is 2.

The equation of the vertical parabola in the vertex form is

(x-h)² = 4a(y-k)

(x-3)² = 4(2)(y-(-1))

** (x-3)² = 8(y+1)**

Simplifying the equation we get

x²-6x+9 = 8y+8

**x²-6x-8y+1=0**

**Related Topics**

Parents and teachers can guide the students to do the problems in this page *Equation of parabola* using the same methods discussed above. If you are having any doubt please contact us, we will help you to clear your doubt.

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