Question

Find the equation of the parabola if  the focus is at (0, 0) and vertex is at the intersection of the lines x + y = 1 and x − y = 3. 

Answer 1

In a parabola, the vertex is the mid-point of the focus and the point of intersection of the axis and the directrix.

Let (x₁, y₁) be the coordinates of the point of intersection of the axis and directrix. 

The point of intersection of \[x + y = 1 \text{ and } x - y = 3\] 

Thus, the vertex and the focus of the parabola are (2, −1) and (0, 0), respectively.

∴ Slope of the axis of the parabola = \[\frac{0 + 1}{0 - 2} = \frac{- 1}{2}\] 

The slope of the directrix is 2.
Let the directrix intersect the axis at K (r, s).

\[\frac{r + 0}{2} = 2, \frac{s + 0}{2} = - 1\]
\[ \Rightarrow r = 4, s = - 2\] 

The required equation of the directrix is

\[y + 2 = 2\left( x - 4 \right)\]which can be rewritten as \[y - 2x + 10 = 0\] 

Let P (x, y) be any point on the parabola whose focus is S (0, 0) and directrix is \[y - 2x + 10 = 0\] 

Draw PM perpendicular to \[y - 2x + 10 = 0\] 

Then, we have:

\[SP = PM\]
\[ \Rightarrow S P^2 = P M^2 \]
\[ \Rightarrow \left( x - 0 \right)^2 + \left( y - 0 \right)^2 = \left( \frac{y - 2x + 10}{\sqrt{5}} \right)^2 \]
\[ \Rightarrow 5 x^2 + 5 y^2 = \left( y - 2x + 10 \right)^2 \]
\[ \Rightarrow x^2 + 4 y^2 + 4xy + 40x - 20y - 100 = 0\]
\[ \Rightarrow \left( x + 2y \right)^2 + 40x - 20y - 100 = 0\]