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FP3 mistake?



a) a2= 9 so a = 3. So I'm assuming the equation for the directix is a mistake (since they have done 9/root5 which is a2/root5 rather than 3/root5 which is what it should be)?
b) the equation for the directrix of b is also a mistake?
ms.png33.9KB
Reply 1
Avatar for Basilla
Basilla
1
a) Note that e=53 e=\frac{\sqrt{5}}{3} - so, 3e=3×35 \frac{3}{e} = 3\times\frac{3}{\sqrt{5}} - that's why its 95\frac{9}{\sqrt{5}}
b) For ellipses, when b>ab>a it's all sort of reversed. For example, instead of b2=a2(1−e2)b^2=a^2(1-e^2), it's a2=b2(1−e2)a^2=b^2(1-e^2).
The focus is at be be instead of ae ae - but, since the focus now lies on the y-axis (it's like you've rotated it 90 degrees), it will be (0,+be) (0, +be) and (0,−be) (0, -be)
Original post by GPODT


a) a2= 9 so a = 3. So I'm assuming the equation for the directix is a mistake (since they have done 9/root5 which is a2/root5 rather than 3/root5 which is what it should be)?
b) the equation for the directrix of b is also a mistake?


How could the directrix be x=3/root5? That would mean it was inside the ellipse.
Reply 3
Avatar for GPODT
GPODT
OP
16
Original post by Basilla
a) Note that e=53 e=\frac{\sqrt{5}}{3} - so, 3e=3×35 \frac{3}{e} = 3\times\frac{3}{\sqrt{5}} - that's why its 95\frac{9}{\sqrt{5}}
b) For ellipses, when b>ab>a it's all sort of reversed. For example, instead of b2=a2(1−e2)b^2=a^2(1-e^2), it's a2=b2(1−e2)a^2=b^2(1-e^2).
The focus is at be be instead of ae ae - but, since the focus now lies on the y-axis (it's like you've rotated it 90 degrees), it will be (0,+be) (0, +be) and (0,−be) (0, -be)


Original post by brianeverit
How could the directrix be x=3/root5? That would mean it was inside the ellipse.


Ah thanks. I understand it now!
in the second sketch the roots should be at 4 & -4 :confused:
Reply 5
Avatar for Basilla
Basilla
1
Original post by the bear
in the second sketch the roots should be at 4 & -4 :confused:

Yeah that looks like an error XD

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