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Q.
The equation of the parabola whose focus is (0, 0) and the tangent at the vertex is x - y + 1 = 0 is
Conic Sections
Solution:
The length of the perpendicular drawn from the given focus upon the given line
$x-y + 1 = 0$ is $\frac{0-0+1}{\sqrt{\left(1\right)^{2}+\left(-1\right)^{2}}} = \frac{1}{\sqrt{2}}.$
The directrix is parallel to the tangent at the vertex.
So, the equation of the dyirecxtrix is $x-y + \lambda = 0$ where $\lambda$ is a constant to be determine.
But the distance between the focus and the directrix $= 2 ×$ (the distance between the focus and the tangent at the vertex)
$= 2\times \frac{1}{\sqrt{2}} = \sqrt{2}.$
Hence $\frac{0-0+\lambda}{\sqrt{\left(1\right)^{2}+\left(-1\right)^{2}}} = \sqrt{2}.$
$\therefore \lambda = 2. \quad$ [$\lambda$ must be positive see figure]
$\therefore $ The directrix is the line $x - y + 2 =0$.
Let $\left(x, y\right)$ be a moving point on the parabola. By the focus-directrixproperty of the parabola, its equation is
$\left(x-0\right)^{2} + \left(y-0\right)^{2} = \left(\pm \frac{x-y+2}{\sqrt{2}}\right)^{2}$
or $x^{2} + y^{2} + 2xy - 4x + 4 y - 4 = 0 .$
