Question

Find the equation of the parabola, if

(i) the focus is at (a, 0) and the vertex is at (a′, 0)

(ii) the focus is at (0, 0) and the vertex is at the intersection of the lines x + y = 1 and x – y = 3.

Answer 1

(i) the focus is at (a, 0) and the vertex is at (a′, 0)

Given:

Focus = (a, 0)

Vertex = (a′, 0)

Let us find the slope of the axis (m₁) = (0-0)/(a′, a)

= 0/(a′, a)

= 0

We know, the products of the slopes of the perpendicular lines is – 1 for non – vertical lines.

Here the slope of the axis is equal to the slope of the x – axis. So, the slope of directrix is equal to the slope of y – axis i.e., ∞.

So, m₂ = ∞

Now let us find the point on directrix.

(a′, 0)= [(x+a/2), (y+0)/2]

By equating we get,

(x+a/2) = a′ and (y)/2 = 0

x + a = 2a′ and y = 0

x = (2a′ – a) and y = 0

So the point on directrix is (2a′ – a, 0).

We know that the equation of the lines passing through (x₁, y₁) and having slope m is y – y₁ = m(x – x₁)

y – (0) = ∞(x – (2a′ – a))

y/∞ = x + a – 2a′

x + a – 2a′ = 0

Now, let us assume P(x, y) be any point on the parabola.

The distance between two points (x₁, y₁) and (x₂, y₂) is given as:

And the perpendicular distance from the point (x₁, y₁) to the line ax + by + c = 0 is 

So by equating both, we get

By cross multiplying we get,

x² + y² – 2ax + a² = x² + a² + 4(a′)² + 2ax – 4aa′ – 4a′x

y² – (4a – 4a′)x + a² – 4(a′)² + 4aa′ = 0

∴ The equation of the parabola is y² – (4a – 4a′)x + a² – 4(a′)² + 4aa′ = 0

(ii) the focus is at (0, 0) and the vertex is at the intersection of the lines x + y = 1 and x – y = 3.

Given:

Focus = (0, 0)

Vertex = intersection of the lines x + y = 1 and x – y = 3.

So the intersecting point of above lines is (2, -1)

Vertex = (2, -1)

Slope of axis (m₁) = (-1-0)/(2-0)

= -1/2

We know that the products of the slopes of the perpendicular lines is – 1.

Let us assume m₂ be the slope of the directrix.

m₁.m₂ = -1

-1/2 . m₂ = -1

So m₂ = 2

Now let us find the point on directrix.

(2, -1) = [(x+0)/2, (y+0)/2]

(x)/2 = 2 and y/2 = -1

x = 4 and y = -2

The point on directrix is (4, – 2).

We know that the equation of the lines passing through (x₁, y₁) and having slope m is y – y₁ = m(x – x₁)

y – (- 2) = 2(x – 4)

y + 2 = 2x – 8

2x – y – 10 = 0

Now, let us assume P(x, y) be any point on the parabola.

The distance between two points (x₁, y₁) and (x₂, y₂) is given as:

And the perpendicular distance from the point (x₁, y₁) to the line ax + by + c = 0 is 

So by equating both, we get

By cross multiplying, we get

5x² + 5y² = 4x² + y² – 40x + 20y – 4xy + 100

x² + 4y² + 4xy + 40x – 20y – 100 = 0

∴ The equation of the parabola is x² + 4y² + 4xy + 40x – 20y – 100 = 0