Question

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

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

Answer 1

The correct option is C $x^2+y^2+2xy-4x+4y-4=0$

The distance between the focus and the tangent at the vertex is $\frac{|0-0+1|}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt{2}}$
The directrix is the line parallel to the tangent at vertex and at a distance $2\times \frac{1}{\sqrt{2}}$ from the focus.
Let the equation of the directirx be,
$x-y+\lambda =0$
So,
$\left|\frac{\lambda }{\sqrt{1^2+1^2}}\right|=\frac{2}{\sqrt{2}}$
$\Rightarrow \lambda =2$
Let $P(x,y)$ be any moving point on the parabola. Then,
$OP=PM$
$x^2+y^2={\left(\frac{x-y+2}{\sqrt{1^2+1^2}}\right)}^2$
$\Rightarrow 2x^2+2y^2=(x-y+2{)}^2$
$\Rightarrow x^2+y^2+2xy-4x+4y-4=0$