Question

The equation of the parabola whose vertex is (a, 0) and the directrix has the equation x + y = 3a, is 

A. x² + y² + 2xy + 6ax + 10ay + 7a² = 0 

B. x² - 2xy + y² + 6ax + 10ay - 7a² = 0 

C. x² - 2xy + y² - 6ax + 10ay - 7a² = 0 

D. none of these

Answer 1

Option : (B)

Given,

The equation of directrix(M) is x + y = 3a and vertex is (a, 0).

We know that,

The directrix and axis are perpendicular to each other. 

The axis also passes through the vertex. 

Let us find the slope of the directrix.

We know that,

The slope of the line ax + by + c = 0 is \(\frac{-a}{b}\).

⇒ m₁ = \(\frac{-1}{1}\)

⇒ m₁ = - 1.

We know that,

The product of the slopes of the perpendicular lines is - 1. 

Let us assume m₂ be the slope of the axis. 

⇒ m₁.m₂ = - 1 

⇒ (-1).m₂ = - 1 

⇒ m₂ = 1 

We know that,

The equation of the line passing through the point (x₁, y₁) and having slope m is (y - y₁) = m(x - x₁)

⇒ y - 0 = 1(x-a)

⇒ y = x - a 

⇒ x - y - a = 0 

On solving the lines x + y = 3a and x - y - a = 0, we get 

The intersection point to be (2a, a). 

We know that,

Vertex is the mid - point of focus and point of intersection of directrix and axis. 

Let (x₁, y₁) be the focus.

⇒ 2a + x₁ = 2a and a + y₁ = 0 

⇒ x₁ = 0 and y₁ = - a. 

The focus is S(0, - a). 

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

We know that,

The point on the parabola is equidistant from focus and directrix. 

We know that,

The distance between two points (x₁, y₁) and (x₂, y₂) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).

We know that,

The perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is \(\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\).

⇒ SP = PM 

⇒ SP² = PM²

⇒ 2x² + 2y² + 4ay + 2a² = x² + y² + 9a² + 2xy - 6ax - 6ay 

⇒ x² + y² - 2xy + 6ax + 10ay - 7a² = 0