Question

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

पर्याय

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

• x² − 2xy + y² + 6ax + 10ay − 7a² = 0 

• x² − 2xy + y² − 6ax + 10ay − 7a² = 0 

• none of these

Answer 1

x² − 2xy + y² + 6ax + 10ay − 7a² = 0

Given:
The vertex is at (a, 0) and the directrix is the line x + y = 3a.
The slope of the line perpendicular to x + y = 3a is 1.
The axis of the parabola is perpendicular to the directrix and passes through the vertex.
∴ Equation of the axis of the parabola = \[y - 0 = 1\left( x - a \right)\]          (1) 

Intersection point of the directrix and the axis is the intersection point of (1) and x + y = 3a.
Let the intersection point be K.
Therefore, the coordinates of K are \[\left( 2a, a \right)\] 

The vertex is the mid-point of the segment joining K and the focus (h, k).
∴ \[a = \frac{2a + h}{2}, 0 = \frac{a + k}{2}\]
\[h = 0, k = - a\] 

Let P (x, y) be any point on the parabola whose focus is S (h, k) and the directrix is x + y= 3a. 

Draw PM perpendicular to x + y = 3a.
Then, we have: \[SP = PM\]
\[ \Rightarrow S P^2 = P M^2 \]
\[ \Rightarrow \left( x - 0 \right)^2 + \left( y + a \right)^2 = \left( \frac{x + y - 3a}{\sqrt{2}} \right)^2 \]
\[ \Rightarrow x^2 + \left( y + a \right)^2 = \left( \frac{x + y - 3a}{\sqrt{2}} \right)^2 \]
\[ \Rightarrow 2 x^2 + 2 y^2 + 2 a^2 + 4ay = x^2 + y^2 + 9 a^2 + 2xy - 6ax - 6ay\]
\[ \Rightarrow x^2 + y^2 - 7 a^2 + 10ay + 6ax - 2xy = 0\]