Question

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

Answer 1

just draw the diagram it will become clear

Answer 2

Hi, Welcome to Open study. Since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. People will still help, so don't worry.

Please consider rewriting your post.

Thanks,

Answer 3

Well I already know the answer just asking to see what the others reply

Answer 4

Directrix is y= -x
Focus is : 2sqrt, 2sqrt
Equation is : (y-sqrt2)^2= 4sqrt2 (x-sqrt2)

Am I right?

Answer 5

you are right with the diretrix but reconsider your thoughts about the equation, its slighty long than you think

Answer 6

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.

Answer 7

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

Answer 8

Okay, but in order to suggest you a better approach we need to know what you have tried first.

Answer 9

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"

Answer 10

Is the equation : \[(x-\sqrt{2}) ^2 + (y-\sqrt{2})^2 = (x+y)^2/2\] ??

Answer 11

Even 2006 won't direct the aspirants to plot the parabola...

Answer 12

err...whatever guys. Can't you just answer the question instead of arguing?

Answer 13

well the equation is
\[\left( x-y \right)^{2}=8(x+y-2)\]

Answer 14

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
\[(x-y)^{2} = 8(x+y-2)\]