question_answer

The equation of the parabola whose focus is (0, 0) and the tangent at the vertex is \[8x-y+1=0\] 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) \[2{{x}^{2}}+2{{y}^{2}}-4xy-x+y-4=0\]

Correct Answer: A

Solution :

[a]

The length of the perpendicular drawn from the given focus upon the given line

\[x-y+1=0\] is \[\frac{0-0+1}{\sqrt{{{(1)}^{2}}+{{(-1)}^{2}}}}=\frac{1}{\sqrt{2}}.\]

The directrix is parallel to the tangent at the vertex.

So, the equation of the directrix 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 \[\frac{0-0+\lambda }{\sqrt{{{(1)}^{2}}+{{(-1)}^{2}}}}=\sqrt{2}.\]

\[\therefore \lambda =2.\] [\[\lambda \] Must be positive see figure]

\[\therefore \] The directrix is the line\[x-y+2=0\].

Let (x, y) be a moving point on the parabola. By The focus-directrix property of the parabola, its equation is

\[{{(x-0)}^{2}}+{{(y-0)}^{2}}={{\left( \pm \frac{x-y+2}{\sqrt{2}} \right)}^{2}}\]

Or \[{{x}^{2}}+{{y}^{2}}+2xy-4x+4y-4=0\]