The axis of a parabola is y=x and it lies in the 1st quadrant, the distance between the origin and the vertex is sqrt(2) its distance between the origin and the focus is 2sqrt(2), what is the equation of the parabola draw the parabola, diretrix and the axis of the parabola and point out its vertex, focus and the foot of the diretrix
just draw the diagram it will become clear
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Well I already know the answer just asking to see what the others reply
Directrix is y= -x Focus is : 2sqrt, 2sqrt Equation is : (y-sqrt2)^2= 4sqrt2 (x-sqrt2) Am I right?
you are right with the diretrix but reconsider your thoughts about the equation, its slighty long than you think
This is not a question-answer site, we emphasize on the understanding of the subject. So if you already have the answer, why don't you post it here? That way we can help you more.
well see this question came in the IITs in 2006 and my working was slighty huge I want to see if there are any time effective methods of solving them
Okay, but in order to suggest you a better approach we need to know what you have tried first.
And I don't think the current form of IIT requires the second part of the question. "draw the parabola, diretrix and the axis of the parabola and point out its vertex, focus and the foot of the diretrix"
Is the equation : \[(x-\sqrt{2}) ^2 + (y-\sqrt{2})^2 = (x+y)^2/2\] ??
Even 2006 won't direct the aspirants to plot the parabola...
err...whatever guys. Can't you just answer the question instead of arguing?
well the equation is \[\left( x-y \right)^{2}=8(x+y-2)\]
One way is to write an equation for parabola on the y-axis, then rotate it 45 degrees \[y = \frac{\sqrt{2}}{8}x^{2} + \sqrt{2}\] \[x' = x \cos(45) - y \sin(45) = \frac{\sqrt{2}}{2}(x-y)\] \[y' = x \sin(45) +y \cos(45) =\frac{\sqrt{2}}{2}(x+y)\] \[\frac{\sqrt{2}}{2}(x+y) = \frac{\sqrt{2}}{8}[\frac{\sqrt{2}}{2}(x-y)]^{2}+\sqrt{2}\] \[x+y = \frac{1}{8}(x-y)^{2} +2\] which is equivalent to \[(x-y)^{2} = 8(x+y-2)\]
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