Question

The equation of the parabola whose focus is (0, 0) and the tangent at the vertex is x - y + 1 = 0 is

• A. $x^2+y^2+2xy-4x+4y-4=0$
• B. $x^2-4x+4y-4=0$
• C. $y^2-4x+4y-4=0$
• D. $2x^2+2y^2-4xy-x+y-4=0$

Answer 1

Answer: (A)

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 $=2\times$ (the distance between the focus and the tangent at the vertex)
$=2\times \frac{1}{\sqrt{2}}=\sqrt{2}.$
Hence $\frac{0-0+\lambda }{\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2}}=\sqrt{2}.$
$\therefore \lambda =2.$ [$\lambda$ 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 $x^2+y^2+2xy-4x+4y-4=0.$