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
3 years, 12 months ago

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The formulas that you used are only valid for parabolas of the form x2=4ay x^2 = 4 ay .When you use a different (orthogoanl) basis like (3x)2=4a(2y+3) (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 aa. Instead, we have to write it as x2=4×(29a)×(y+32) x^2 = 4 \times \left( \frac{2}{9} a \right) \times ( y + \frac{3}{2} ) , to conclude that the distance should be 29a \frac{2}{9} a instead.

In this case, you did a change of basis coordinates from (x,y) (x,y) to (xy,x+y+4) (x-y, x+y + 4) . This changed your distance by a factor of 2 \sqrt{2} , and so the true distance would have been 82=42 \frac{8}{ \sqrt{2} } = 4 \sqrt{2} . This agrees with the calculation that 42+42=42 \sqrt{ 4 ^2 + 4^2 } = 4 \sqrt{2} .

Calvin Lin Staff - 3 years, 12 months ago

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Thank you.I understand my mistake

Hari prasad Varadarajan - 3 years, 12 months ago

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