ATAR Notes: Forum
HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Advanced => Topic started by: _OwO_ on November 06, 2018, 09:36:31 pm
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Hey! please send help
how would you do this question:
find the equation of the parabola with the coordinates of the vertex (0,0) and equation of the axis x=0 and passing through the point (-8,2)
thankyou!
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Hey there! Super sorry for the late reply my friend, sort of stopped looking at the maths boards for a bit in the HSC rush - Keen to give you a hand though ;D
So the first two bits of info are key here. We have a parabola with a vertex at \((0,0)\) and an axis of \(x=0\) (the y-axis). This means we have a parabola that is vertical, with its turning point at the origin (just a really plain, vanilla parabola really!). Any parabola on the origin like this takes the form:

To find the actual equation, we need \(a\). To get it, substitute the point we are given:
\implies (-8)^2=4a(2)\\64=8a\\a=8)
So substituting that in, we get \(x^2=32y\), or \(y=\frac{x^2}{32}\). That is the equation we need! I hope this helps ;D