Question

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

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

Answer 1

Answer: (C)

$x^2+y^2+2xy-4x+4y-4=0$

Focus is at $(0,0)$

Tangent at the vertex is

$x-y+1=0$ or $y=x+1$ .....(1)

Equation of normal to the tangent will be the axis of parabola

$y-0=-1(x-0)$ or $y+x=0$ .....(2)

Solving (1) and (2), we get $x=-\frac{1}{2},y=\frac{1}{2}$

So, the coordinates of the vertex is $A(-\frac{1}{2},\frac{1}{2})$.

Let $Z(p,q)$ be the point on the directrix.

$\frac{p+0}{2}=-\frac{1}{2}$ and $\frac{q+0}{2}=\frac{1}{2}$

$\Rightarrow p=-1,q=1$

So, the coordinates of $Z$ are $(-1,1)$.

Equation of directrix will be of the form

$x-y+k=0$

As it passes through $Z(-1,1)$, we get $k=2$

So, the equation of directrix is $x-y+2=0$

Let $M$ be a point on the directrix such that

$OP=PM$

$O{P}^2=P{M}^2$

$x^2+y^2=\frac{(x-y+2{)}^2}{2}$

$\Rightarrow x^2+y^2+2xy-4x+4y-4=0$