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# Where is the mistake

Hi guys.. I was just solving a problem related to parabolas and came across this situation.Everything I did seems right but there seems to be a mistake somewhere. Here goes the situation

There is a parabola (x-y)^2 = -48(x+y+4). So I figured that the axis must be x-y=0 and the tangent at the vertex must be x+y+4=0.So the vertex should be (-2,-2). Using the value of 'a' that is 8, I figured that the directrix should be 8 units away from the tangent at the vertex(this is because if you tilt the axes so that the axis of the parabola becomes the y axis,the distance between the tangent at the vertex and directrix must be 8.So even if I again tilt the axes back to the original position the distance between them remains the same) and found it to be x+y-4=0.So the point where the directrix intersects the axis is (2,2). Now comes the interesting part.When I found the distance between (2,2) and(-2.-2) it was not 8.I am shocked. I found the directrix taking the distance to be 8.But why isn't the distance not 8?

Note by Hari prasad Varadarajan
1 year, 6 months ago

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The formulas that you used are only valid for parabolas of the form $$x^2 = 4 ay$$.When you use a different (orthogoanl) basis like $$(3x)^2 = 4 a ( 2y + 3 )$$ then you have to consider how the diagram scales accordingly. In this example, note that the distance isn't just the value of $$a$$. Instead, we have to write it as $$x^2 = 4 \times \left( \frac{2}{9} a \right) \times ( y + \frac{3}{2} )$$, to conclude that the distance should be $$\frac{2}{9} a$$ instead.

In this case, you did a change of basis coordinates from $$(x,y)$$ to $$(x-y, x+y + 4)$$. This changed your distance by a factor of $$\sqrt{2}$$, and so the true distance would have been $$\frac{8}{ \sqrt{2} } = 4 \sqrt{2}$$. This agrees with the calculation that $$\sqrt{ 4 ^2 + 4^2 } = 4 \sqrt{2}$$. Staff · 1 year, 6 months ago

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Thank you.I understand my mistake · 1 year, 6 months ago

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