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